Integrated Mathematics 1

Course overview

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Integrated Mathematics 1 braids algebra, functions, statistics and geometry into one year instead of teaching them as separate courses, which is exactly what makes it hard to revise from a textbook, the coordinate geometry in Unit 6 leans on the slope work in Unit 2, and the exponential models in Unit 4 only make sense against the linear ones before them. The topics below follow the order most California Integrated I courses use, so you can go straight to whatever was covered in class today.

U1Unit 1: Expressions, Equations and Units4 topics
U2Unit 2: Linear Functions4 topics
U3Unit 3: Systems of Equations and Inequalities3 topics
U4Unit 4: Sequences and Exponential Functions4 topics
U5Unit 5: Descriptive Statistics3 topics
U6Unit 6: Transformations, Congruence and Coordinates4 topics

All six units are open, 22 topics in all. Every topic opens with the method, worked examples, and fifteen practice problems. Every problem has a full worked solution, so you can find the step where yours went wrong. Each unit closes with a ten-problem mixed review.

Free preview: open any 5 topics without an account. The counter on the left keeps track.

Topic 1.1 · Unit 1 · CA CCSS N-Q.1–3

Quantities, units and unit conversion

The arithmetic here is easy and the setup is where the marks go. Write every conversion as a fraction and let the units cancel, rather than deciding in your head whether to multiply or divide.

The method

Write what you have as a fraction, even when it looks like one number. 45 miles per hour is $ \frac{45 \text{ mi}}{1 \text{ hr}} $.
Write each conversion as a fraction equal to 1. Since $ 1 \text{ mi} = 5280 \text{ ft} $, both $ \frac{5280 \text{ ft}}{1 \text{ mi}} $ and $ \frac{1 \text{ mi}}{5280 \text{ ft}} $ equal 1, so multiplying by either changes the units and never the value.
Orient it so the unit you want gone cancels. If "mi" is on top of what you have, put "mi" on the bottom of the conversion.
Cancel, then multiply. Whatever units survive are your answer's units. If they are not the ones asked for, a fraction is upside down.
Check the size. Converting to a smaller unit gives a bigger number.

Where marks are lost: squared and cubed units. To convert $ \text{ft}^2 $ to $ \text{in}^2 $ the conversion must be squared too, $ \left(\frac{12 \text{ in}}{1 \text{ ft}}\right)^2 = \frac{144 \text{ in}^2}{1 \text{ ft}^2} $, not 12. Using 12 gives an answer twelve times too small.

Worked examples

Example 1: a rate with both units changing. Express 60 miles per hour in feet per second.

Two units change, so two conversion fractions. Miles cancels from the top, hours from the bottom:

$ \frac{60 \text{ mi}}{1 \text{ hr}} \times \frac{5280 \text{ ft}}{1 \text{ mi}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = \frac{60 \times 5280}{3600} = 88 \text{ ft/s} $

Only ft over s survives, which is what was asked. Sanity check: 88 is large for a modest speed, which is right, because feet and seconds are both small units.

Example 2: a squared unit. Convert 12 square meters to square centimeters.

$ 12 \text{ m}^2 \times \left(\frac{100 \text{ cm}}{1 \text{ m}}\right)^2 = 12 \times 10000 = 120{,}000 \text{ cm}^2 $

Multiplying by 100 instead of 10 000 is the standard error and lands a hundredfold short.

Example 3: precision. A recipe for 4 uses 350 g of flour. How much for 7?

$ 350 \times \frac{7}{4} = 612.5 \text{ g} $. But the original was given to the nearest 10 g, so 612.5 claims precision the source never had. State it as about 610 g, an answer should not be more precise than the data behind it.

Practice · 15 problems

Work each one on paper first. Every solution shows the full setup, so a wrong answer can be traced to the line that caused it.

Convert 5 kilometers to meters.
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$ 5 \text{ km} \times \frac{1000 \text{ m}}{1 \text{ km}} = 5000 \text{ m} $. Meters are smaller, so the number grows. 5000 m
Convert 3 hours to seconds.
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$ 3 \times 60 \times 60 = 10800 $. 10 800 s
A runner covers 400 meters in 50 seconds. Find the speed in meters per second.
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$ \frac{400 \text{ m}}{50 \text{ s}} = 8 \text{ m/s} $. 8 m/s
Convert 90 kilometers per hour to meters per second.
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$ \frac{90 \text{ km}}{1 \text{ hr}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = \frac{90000}{3600} = 25 $. 25 m/s
Convert 4 square meters to square centimeters.
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Square the conversion: $ 4 \times 100^2 = 4 \times 10000 = 40000 $. 40 000 cm²
Convert 2 cubic feet to cubic inches.
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Cube it: $ 2 \times 12^3 = 2 \times 1728 = 3456 $. 3456 in³
Petrol costs $3.60 per gallon. Express this in dollars per liter, given 1 gal = 3.785 L.
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Only the bottom unit changes, so gallons goes on top of the conversion to cancel: $ \frac{\$3.60}{1 \text{ gal}} \times \frac{1 \text{ gal}}{3.785 \text{ L}} \approx \$0.95 $. A liter is smaller than a gallon, so the price per liter must be smaller, a useful check. about $0.95/L
A field is 120 m by 75 m. Find its area in hectares (1 ha = 10 000 m²).
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Area first: $ 120 \times 75 = 9000 \text{ m}^2 $. Then $ 9000 \times \frac{1}{10000} = 0.9 $. 0.9 ha
In $ d = rt $, $ r $ is in miles per hour and $ t $ in hours. What units does $ d $ have, and why?
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$ \frac{\text{mi}}{\text{hr}} \times \text{hr} $, the hours cancel and miles survives. Worth doing on formulas as well as conversions: if the surviving units are not a distance, the formula has been used wrongly. miles
A car uses 8 liters per 100 km. How much fuel for a 250 km journey?
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$ \frac{8 \text{ L}}{100 \text{ km}} \times 250 \text{ km} = 20 $. The km cancels. 20 L
A printer produces 18 pages per minute. How long, in hours, to print 5400 pages?
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Invert the rate so pages cancel: $ 5400 \times \frac{1 \text{ min}}{18} = 300 \text{ min} $, then $ 300 \div 60 = 5 $. 5 hours
A scale drawing uses 1 cm for 2.5 m. A wall measures 6.4 cm on the drawing. How long is the real wall?
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The scale is a conversion fraction like any other: $ 6.4 \times 2.5 = 16 $. 16 m
Water flows at 2.5 cubic meters per minute. Express this in liters per second (1 m³ = 1000 L).
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$ \frac{2.5 \text{ m}^3}{1 \text{ min}} \times \frac{1000 \text{ L}}{1 \text{ m}^3} \times \frac{1 \text{ min}}{60 \text{ s}} = \frac{2500}{60} \approx 41.7 $. about 41.7 L/s
A student converts 5 m² to cm² and gets 500 cm². Find the error and give the correct answer.
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They multiplied by 100 once, treating m² as m. The unit is squared so the conversion must be squared: $ 5 \times 100^2 = 50000 $. Their answer is a hundred times too small. 50 000 cm²
A time is measured as 3.0 seconds. A student computes $ 100 \div 3.0 = 33.33333 $ m/s and writes every digit. Why is that wrong, and what should they write?
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The time was measured to two significant figures, so the result cannot be known to seven, the extra digits claim precision the measurement never had. Round to match the least precise input. 33 m/s

Topic 1.2 · Unit 1 · CA CCSS A-SSE.1–2

Reading and building expressions

Two skills share this topic. One is mechanical, simplify correctly. The other is interpretive, say what a particular part of an expression means in the situation it describes. This course tests the second more than students expect, and it cannot be pattern-matched.

The method

Vocabulary you are expected to use exactly. In $ 5x^2 + 3x - 7 $: the terms are $ 5x^2 $, $ 3x $, $ -7 $; the coefficient of $ x^2 $ is 5; $ -7 $ is the constant. A factor is something multiplied, so $ 3(x+2) $ has factors 3 and $ (x+2) $.

To simplify: distribute across every bracket, remembering that a minus in front multiplies every term inside. Then combine like terms, identical variable parts only. $ 3x $ and $ 3x^2 $ are not like terms.
To interpret a part in context: ask what the whole expression counts, then what changes when that part changes. A number attached to a variable is a rate; a number standing alone is a starting value or one-off.
To test equivalence: simplify both and compare. One value agreeing is not proof; one value disagreeing is proof they differ.

Where marks are lost: the negative outside a bracket. $ 8 - 2(x - 3) $ is $ 8 - 2x + 6 = 14 - 2x $, not $ 8 - 2x - 6 $. The $ -2 $ multiplies the $ -3 $ as well, and two negatives make a plus.

Worked examples

Example 1: a negative outside a bracket. Simplify $ 4(2x + 5) - 3(x - 4) $.

$ 4(2x+5) = 8x + 20 $; $ -3(x-4) = -3x + 12 $. Note the $ +12 $, since $ -3 \times -4 = +12 $. Combining: $ 5x + 32 $.

Example 2: interpreting in context. A gym costs $ C = 45 + 28m $ after $ m $ months. What do 45 and 28 represent?

Ask what changes with $ m $. The 28 multiplies the months, so it is the cost per month, $28/month. The 45 is paid whatever $ m $ is, including $ m = 0 $, so it is the one-off joining fee. The giveaway is always whether the number is attached to the variable.

Example 3: reading a factored form. Weekly profit is $ P = 15(n - 40) $, where $ n $ is items sold. What does the 40 tell you?

Profit is zero when the bracket is zero, at $ n = 40 $, the break-even point. Below it the bracket is negative and the shop loses money. Multiplied out as $ 15n - 600 $ that fact is hidden, which is the whole reason the factored form is useful.

Practice · 15 problems

1–6 are direct practice, 7–12 mix the skills, and 13–15 ask you to interpret or diagnose rather than simplify.

Simplify $ -2(3x - 5) $.
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$ -2 \times 3x = -6x $; $ -2 \times -5 = +10 $. $ -6x + 10 $
Simplify $ 7x + 2y - 3x + 5y $.
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Only matching variable parts combine: $ 7x - 3x = 4x $, $ 2y + 5y = 7y $. $ 4x + 7y $
Simplify $ 8 - 2(x - 3) $.
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$ -2(x-3) = -2x + 6 $, then $ 8 + 6 = 14 $. $ 14 - 2x $. Writing $ 8 - 2x - 6 $ is the standard error.
Simplify $ 3x^2 + 2x - x^2 + 5x $.
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$ x^2 $ terms: $ 3 - 1 = 2 $. $ x $ terms: $ 2 + 5 = 7 $. They cannot combine with each other. $ 2x^2 + 7x $
Simplify $ \frac{1}{2}(6x - 8) + 3 $.
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$ 3x - 4 $, then $ -4 + 3 = -1 $. $ 3x - 1 $
Simplify $ 2(3x - 1) - (x + 5) $.
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A bare minus before a bracket is $ -1 \times $ everything inside: $ -(x+5) = -x - 5 $. With $ 6x - 2 $: $ 5x - 7 $
In $ 7x^2 - 4x + 9 $, name the coefficient of $ x $ and the constant term.
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The coefficient of $ x $ is $ -4 $, the sign belongs to it. The constant is 9. $ -4 $ and 9
Are $ 3(x + 4) $ and $ 3x + 4 $ equivalent? Justify.
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No. $ 3(x+4) = 3x + 12 $. One counterexample settles it: at $ x = 0 $ the first gives 12 and the second gives 4. Not equivalent
Write an expression for the perimeter of a rectangle of width $ w $ whose length is 3 more than twice the width.
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Length $ = 2w + 3 $. Perimeter $ = 2w + 2(2w+3) = 2w + 4w + 6 $. $ 6w + 6 $
Simplify $ 3(2x - 5) - 4(x - 2) + 7 $.
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$ 6x - 15 $, $ -4x + 8 $, $ +7 $. Then $ 6x - 4x = 2x $ and $ -15 + 8 + 7 = 0 $. $ 2x $
A taxi charges $4 plus $2.50 per mile. Write an expression for an $ m $-mile trip.
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The per-mile rate attaches to the variable; the flag-fall stands alone. $ 4 + 2.5m $
Show that $ 2(x + 3) + 4x $ and $ 6(x + 1) $ are equivalent.
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Left: $ 2x + 6 + 4x = 6x + 6 $. Right: $ 6x + 6 $. Identical after simplifying, so equivalent for every $ x $. Both give $ 6x + 6 $
A phone plan costs $ C = 20 + 0.05t $ dollars for $ t $ minutes. Interpret both numbers, with units.
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0.05 attaches to $ t $, so it is 5 cents per minute. 20 is paid even at $ t = 0 $, so it is the fixed monthly fee. $20 fixed; $0.05 per minute
Revenue is $ R = 12(s - 30) $ for $ s $ items sold. What is significant about $ s = 30 $, and why is this form more useful than $ 12s - 360 $?
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At $ s = 30 $ the bracket is zero so $ R = 0 $, break-even. The factored form shows it directly; the expanded form requires solving to find it. Break-even at 30 items
A student simplifies $ 5 - 3(2x - 4) $ to $ 5 - 6x - 12 = -6x - 7 $. Find the error.
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They distributed $ -3 $ to the $ 2x $ correctly but kept the sign on the $ -4 $. $ -3 \times -4 = +12 $, so it is $ 5 - 6x + 12 $. $ 17 - 6x $

Topic 1.3 · Unit 1 · CA CCSS A-REI.1, A-REI.3, A-CED.4

Solving linear equations and rearranging formulas

Solving for $ x $ and rearranging a formula for one of its letters are the same operation, undo what is being done to the target, in reverse order, doing it to both sides. Treating them as two separate skills is why the formula questions feel harder than they are.

The method

Clear fractions first by multiplying every term by the common denominator. Every term, including ones without a fraction.
Distribute across any brackets.
Collect the target on one side and everything else on the other. Move the smaller coefficient to avoid negatives if you can choose.
Factor out the target if it appears in more than one term; this is the step formulas need and simple equations do not. From $ ax + bx = c $ you get $ x(a+b) = c $, so $ x = \frac{c}{a+b} $.
Divide by the whole coefficient and check by substituting back.

Special cases worth recognizing instantly: if the variables cancel and you are left with something true, such as $ 6 = 6 $, every number works, infinitely many solutions. If you are left with something false, such as $ 6 = 9 $, no number works, no solution. Students often write "0" for the second; the answer is that there is no solution, which is not the same as $ x = 0 $.

Worked examples

Example 1: variables on both sides with fractions. Solve $ \frac{x}{3} + 2 = \frac{x}{2} - 1 $.

Common denominator 6, multiply every term: $ 2x + 12 = 3x - 6 $. Then $ 12 + 6 = 3x - 2x $, so $ x = 18 $.

Check: $ \frac{18}{3} + 2 = 8 $ and $ \frac{18}{2} - 1 = 8 $. ✓

Example 2: no solution against infinitely many. Solve $ 4(x - 2) = 4x + 5 $.

$ 4x - 8 = 4x + 5 $. Subtracting $ 4x $ from both sides gives $ -8 = 5 $, which is false. The variable has vanished and left a false statement, so no solution. Had it left $ -8 = -8 $, the answer would be all real numbers.

Example 3: a formula where the target appears twice. Make $ x $ the subject of $ ax + b = cx + d $.

Collect the $ x $ terms: $ ax - cx = d - b $. Now factor, which is the step that makes this solvable: $ x(a - c) = d - b $, so $ x = \frac{d - b}{a - c} $. Without factoring there is no way to isolate a letter that appears in two terms.

Practice · 15 problems

1–6 are equations, 7–11 mix in fractions and special cases, 12–15 are formula rearrangements and word problems.

Solve $ 5x - 7 = 23 $.
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$ 5x = 30 $, so $ x = 6 $. $ x = 6 $
Solve $ 3x + 4 = x + 16 $.
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$ 3x - x = 16 - 4 $, so $ 2x = 12 $ and $ x = 6 $. $ x = 6 $
Solve $ 2(x + 5) = 18 $.
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Distribute: $ 2x + 10 = 18 $, so $ 2x = 8 $ and $ x = 4 $. Dividing both sides by 2 first is quicker and equally valid. $ x = 4 $
Solve $ 7 - 3x = 22 $.
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$ -3x = 15 $, so $ x = -5 $. Dividing by a negative flips the sign of the answer. $ x = -5 $
Solve $ 4(2x - 1) = 3(x + 7) $.
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$ 8x - 4 = 3x + 21 $, so $ 5x = 25 $ and $ x = 5 $. $ x = 5 $
Solve $ \frac{x}{4} = 9 $.
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Multiply both sides by 4: $ x = 36 $. $ x = 36 $
Solve $ \frac{2x}{3} - 1 = \frac{x}{2} + 2 $.
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Multiply every term by 6: $ 4x - 6 = 3x + 12 $, so $ x = 18 $. Forgetting to multiply the $ -1 $ and $ +2 $ is the usual slip. $ x = 18 $
Solve $ 3(x - 2) = 3x - 6 $.
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$ 3x - 6 = 3x - 6 $. The variable cancels leaving a true statement, so every real number satisfies it. Infinitely many solutions
Solve $ 2x + 5 = 2x - 3 $.
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Subtracting $ 2x $ gives $ 5 = -3 $, which is false. No solution: not $ x = 0 $.
Solve $ 0.5x + 1.2 = 3.7 $.
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$ 0.5x = 2.5 $, so $ x = 5 $. Multiplying through by 10 first to clear decimals works equally well. $ x = 5 $
Solve $ \frac{x + 3}{5} = 4 $.
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Multiply both sides by 5: $ x + 3 = 20 $, so $ x = 17 $. The whole numerator is being divided, so the whole numerator gets multiplied. $ x = 17 $
Make $ h $ the subject of $ A = \frac{1}{2}bh $.
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Multiply by 2: $ 2A = bh $. Divide by $ b $: $ h = \frac{2A}{b} $. $ h = \frac{2A}{b} $
Make $ C $ the subject of $ F = \frac{9}{5}C + 32 $.
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$ F - 32 = \frac{9}{5}C $, then multiply by $ \frac{5}{9} $: $ C = \frac{5}{9}(F - 32) $. $ C = \frac{5}{9}(F - 32) $
Make $ x $ the subject of $ mx + n = px - q $.
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Collect: $ mx - px = -q - n $. Factor, the step this type needs: $ x(m - p) = -(q + n) $, so $ x = \frac{-(q+n)}{m-p} $, which can also be written $ \frac{q+n}{p-m} $. $ x = \frac{q+n}{p-m} $
A plumber charges a $60 call-out plus $45 per hour. A job cost $217.50. How many hours did it take?
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Set up $ 60 + 45h = 217.50 $. Then $ 45h = 157.50 $ and $ h = 3.5 $. 3.5 hours

Topic 1.4 · Unit 1 · CA CCSS A-REI.3

Linear inequalities in one variable

Solve an inequality exactly as you would the matching equation, with one extra rule that accounts for most of the lost marks on this topic: multiplying or dividing both sides by a negative reverses the inequality sign.

The method

Solve as an equation: clear fractions, distribute, collect, divide.
Flip the sign if and only if you multiply or divide both sides by a negative number. Adding or subtracting a negative does not flip it.
Graph it. Open circle for $ < $ or $ > $, closed circle for $ \le $ or $ \ge $. Shade toward the numbers that work.
Test one value from the shaded region in the original inequality. This catches a flipped sign immediately and takes ten seconds.
Compound inequalities: "and" means the overlap of the two regions; "or" means both regions together. $ -3 < x \le 5 $ is an "and" written compactly.

Where marks are lost: $ -2x > 8 $ gives $ x < -4 $, not $ x > -4 $. Test it: $ x = -5 $ gives $ -2(-5) = 10 > 8 $ ✓, while $ x = 0 $ gives $ 0 > 8 $ ✗. Avoid the rule entirely by moving the variable to whichever side keeps it positive.

Worked examples

Example 1: the flip. Solve $ 5 - 3x \ge 17 $.

$ -3x \ge 12 $. Dividing by $ -3 $ reverses the sign: $ x \le -4 $.

Check with $ x = -5 $: $ 5 - 3(-5) = 20 \ge 17 $ ✓. Check with $ x = 0 $: $ 5 \ge 17 $ ✗. The solution set is correct.

Example 2: avoiding the flip. Solve $ 4 - 2x < 10 $ by moving the variable instead.

Add $ 2x $ to both sides: $ 4 < 10 + 2x $. Subtract 10: $ -6 < 2x $. Divide by positive 2: $ -3 < x $, that is $ x > -3 $. No flip was ever needed, because the variable was never divided by a negative.

Example 3: a compound inequality. Solve $ -7 < 2x + 1 \le 9 $.

Operate on all three parts at once. Subtract 1: $ -8 < 2x \le 8 $. Divide by 2: $ -4 < x \le 4 $. Graph with an open circle at $ -4 $, a closed circle at 4, and shading between.

Practice · 15 problems

1–6 are direct, 7–11 involve the flip or fractions, and 12–15 are compound inequalities and word problems.

Solve $ x + 7 < 12 $.
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Subtract 7. No multiplication by a negative, so no flip. $ x < 5 $
Solve $ 3x \ge 21 $.
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Divide by positive 3. $ x \ge 7 $
Solve $ 2x - 5 > 9 $.
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$ 2x > 14 $, so $ x > 7 $. $ x > 7 $
Solve $ -4x < 20 $.
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Dividing by $ -4 $ flips the sign: $ x > -5 $. Test $ x = 0 $: $ 0 < 20 $ ✓. $ x > -5 $
Solve $ 6 - x \le 2 $.
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$ -x \le -4 $; dividing by $ -1 $ flips: $ x \ge 4 $. Or move the $ x $ across instead: $ 6 \le 2 + x $, giving $ 4 \le x $. $ x \ge 4 $
Graph the solution to $ x \ge -2 $ on a number line.
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A closed circle at $ -2 $, because $ \ge $ includes the endpoint, with shading to the right. Closed circle at −2, shaded right
Solve $ 5 - 2x > 13 $.
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$ -2x > 8 $, dividing by $ -2 $ flips: $ x < -4 $. Test $ x = -5 $: $ 5 + 10 = 15 > 13 $ ✓. $ x < -4 $
Solve $ 3(x - 2) \le 2x + 1 $.
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$ 3x - 6 \le 2x + 1 $, so $ x \le 7 $. No flip, nothing was divided by a negative. $ x \le 7 $
Solve $ \frac{x}{3} + 2 > 5 $.
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$ \frac{x}{3} > 3 $, multiply by positive 3: $ x > 9 $. $ x > 9 $
Solve $ \frac{2 - x}{4} \ge 1 $.
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Multiply by 4: $ 2 - x \ge 4 $, so $ -x \ge 2 $ and dividing by $ -1 $ flips to $ x \le -2 $. $ x \le -2 $
A student solves $ -3x < 9 $ and writes $ x < -3 $. Find the error.
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They divided by $ -3 $ without reversing the sign. It should be $ x > -3 $. Testing $ x = 0 $ exposes it at once: $ 0 < 9 $ is true, so 0 must be in the solution set, and $ x < -3 $ excludes it. $ x > -3 $
Solve $ -1 \le 3x + 2 < 11 $.
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Subtract 2 from all three parts: $ -3 \le 3x < 9 $. Divide all by 3: $ -1 \le x < 3 $. $ -1 \le x < 3 $
Solve $ x - 1 < 3 $ or $ x + 2 > 10 $.
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Separately: $ x < 4 $ or $ x > 8 $. "Or" keeps both regions, and they do not overlap, so the solution is two separate pieces. $ x < 4 $ or $ x > 8 $
A lift holds at most 600 kg. Four people weighing a total of 290 kg are inside. Boxes weigh 25 kg each. How many boxes can be loaded?
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$ 290 + 25b \le 600 $, so $ 25b \le 310 $ and $ b \le 12.4 $. Boxes are whole, so at most 12. Rounding 12.4 up to 13 would exceed the limit, context decides the rounding direction. 12 boxes
A student needs a mean of at least 80 across four tests. They have scored 76, 84 and 79. What must they score on the fourth?
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$ \frac{76 + 84 + 79 + x}{4} \ge 80 $, so $ 239 + x \ge 320 $ and $ x \ge 81 $. At least 81

Unit 1 mixed review · 10 problems · all topics

Unit 1 mixed review: Expressions, Equations and Units

These are shuffled across all four topics and do not tell you which method they want, which is what makes them closer to a real test than a single topic's practice set.

Convert 72 kilometers per hour to meters per second.
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$ \frac{72 \text{ km}}{1 \text{ hr}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ hr}}{3600 \text{ s}} = \frac{72000}{3600} = 20 $. 20 m/s
Simplify $ 5(2x - 3) - 2(x - 7) $.
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$ 10x - 15 $ and $ -2x + 14 $. Combining: $ 8x - 1 $. $ 8x - 1 $
Solve $ 4x - 9 = 2x + 7 $.
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$ 2x = 16 $, so $ x = 8 $. $ x = 8 $
Solve $ 3 - 2x \ge 11 $.
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$ -2x \ge 8 $; dividing by $ -2 $ flips the sign: $ x \le -4 $. Test $ x = -5 $: $ 3 + 10 = 13 \ge 11 $ ✓. $ x \le -4 $
Make $ r $ the subject of $ C = 2\pi r $.
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Divide both sides by $ 2\pi $. $ r = \frac{C}{2\pi} $
Convert 3 square meters to square centimeters.
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Square the conversion: $ 3 \times 100^2 = 30000 $. 30 000 cm²
A car rental costs $ C = 45 + 0.28m $ dollars for $ m $ miles. Interpret both numbers.
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0.28 is attached to $ m $, so it is the charge per mile, 28 cents. 45 is paid at $ m = 0 $, so it is the fixed hire fee. $45 fixed; $0.28 per mile
Solve $ 2(x + 4) = 2x + 8 $.
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$ 2x + 8 = 2x + 8 $. The variable cancels leaving a true statement. Infinitely many solutions
Make $ x $ the subject of $ y = mx + b $.
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$ y - b = mx $, then divide by $ m $. $ x = \frac{y - b}{m} $
A delivery van can carry 900 kg. It already holds 340 kg. Crates weigh 35 kg. How many more can it take?
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$ 340 + 35c \le 900 $, so $ 35c \le 560 $ and $ c \le 16 $. Exactly 16 fits. 16 crates

Topic 2.1 · Unit 2 · CA CCSS F-IF.1–2, F-IF.5

Function notation, domain and range

Almost every error on this topic comes from one confusion: $ f(3) $ and "solve $ f(x) = 3 $" are opposite questions. The first hands you an input and asks for the output; the second hands you an output and asks for the input.

The method

To evaluate $ f(3) $: replace every $ x $ in the rule with 3 and compute. The answer is a $ y $-value.
To solve $ f(x) = 3 $: set the rule equal to 3 and solve for $ x $. The answer is an $ x $-value, and there may be more than one.
Domain is every input allowed; range is every output produced. On a graph, domain is how far it extends left to right, range how far up and down.
To decide whether a relation is a function: each input may have only one output. On a graph that is the vertical line test; in a table, look for a repeated $ x $ with different $ y $.
Restrictions on domain come from two places in this course: division by zero, and context. If $ x $ counts people, the domain is whole numbers $ \ge 0 $ however the algebra behaves.

Where marks are lost: $ f(x) $ is not $ f $ multiplied by $ x $; it is one symbol naming an output. And when substituting a negative, use brackets: for $ f(x) = x^2 $, $ f(-3) = (-3)^2 = 9 $, not $ -3^2 = -9 $.

Worked examples

Example 1: evaluating with a negative input. For $ f(x) = 3x^2 - 5x $, find $ f(-2) $.

Substitute with brackets: $ f(-2) = 3(-2)^2 - 5(-2) $. Now $ (-2)^2 = 4 $, so $ 3 \times 4 = 12 $, and $ -5 \times -2 = +10 $. Total $ f(-2) = 22 $. Dropping the brackets and writing $ 3 \times -4 $ is the usual slip.

Example 2: the reverse question. For $ f(x) = 2x + 1 $, solve $ f(x) = 9 $.

This asks which input produces 9, so set the rule equal to 9: $ 2x + 1 = 9 $, giving $ x = 4 $. Compare with $ f(9) = 2(9) + 1 = 19 $, a completely different question with a completely different answer.

Example 3: domain and range in context. A ticket costs $12. Total cost for $ n $ tickets is $ C(n) = 12n $. State the domain and range.

Algebraically $ n $ could be any real number, but you cannot buy 2.5 tickets or −3 tickets. The domain is whole numbers $ n \ge 0 $, and the range is the matching outputs $ \{0, 12, 24, 36, \dots\} $. Context overrides what the algebra alone would allow, and examiners look for exactly this.

Practice · 15 problems

1–6 are direct, 7–12 mix evaluating with solving, and 13–15 involve context and diagnosis.

For $ f(x) = 4x - 7 $, find $ f(3) $.
Show the full solution

$ 4(3) - 7 = 12 - 7 = 5 $. $ f(3) = 5 $
For $ g(x) = x^2 + 1 $, find $ g(-4) $.
Show the full solution

$ (-4)^2 = 16 $, so $ 16 + 1 = 17 $. Brackets matter here. $ g(-4) = 17 $
For $ f(x) = 2x + 5 $, solve $ f(x) = 17 $.
Show the full solution

$ 2x + 5 = 17 $, so $ 2x = 12 $ and $ x = 6 $. $ x = 6 $
For $ h(x) = 3 - x $, find $ h(0) $ and $ h(10) $.
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$ h(0) = 3 $; $ h(10) = 3 - 10 = -7 $. 3 and $ -7 $
Does the table represent a function? $ (1, 4), (2, 7), (1, 9), (3, 2) $
Show the full solution

The input 1 appears twice with different outputs, 4 and 9. An input may have only one output. Not a function
For $ f(x) = 5x $, find $ f(2) + f(3) $.
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$ f(2) = 10 $, $ f(3) = 15 $, so the sum is 25. Note this is not $ f(5) $, though here they happen to agree; that coincidence fails for most functions. 25
For $ f(x) = x^2 - 9 $, solve $ f(x) = 0 $.
Show the full solution

$ x^2 - 9 = 0 $, so $ x^2 = 9 $ and $ x = 3 $ or $ x = -3 $. Both are valid inputs; giving only the positive loses a mark. $ x = \pm 3 $
For $ f(x) = 3x + 2 $, find $ f(a + 1) $.
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Replace every $ x $ with the whole expression $ a+1 $: $ 3(a+1) + 2 = 3a + 3 + 2 $. $ 3a + 5 $
A graph runs from $ x = -2 $ to $ x = 6 $, with $ y $-values from $ -1 $ up to 5. State the domain and range.
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Domain is the horizontal extent, range the vertical. Domain $ -2 \le x \le 6 $; range $ -1 \le y \le 5 $
For $ f(x) = \frac{6}{x - 2} $, what value must be excluded from the domain, and why?
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The denominator cannot be zero: $ x - 2 = 0 $ at $ x = 2 $. Every other real number is allowed. $ x \neq 2 $
For $ f(x) = 2x - 3 $, find $ f(5) $ and solve $ f(x) = 5 $. Explain why the answers differ.
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$ f(5) = 2(5) - 3 = 7 $. Solving $ 2x - 3 = 5 $ gives $ x = 4 $. The first asks what comes out when 5 goes in; the second asks what must go in for 5 to come out. $ f(5) = 7 $; $ x = 4 $
Does $ x = y^2 $ define $ y $ as a function of $ x $? Justify.
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No. At $ x = 4 $, $ y $ could be 2 or $ -2 $, so one input gives two outputs. Its graph is a sideways parabola and fails the vertical line test. Not a function
A tank holds 500 liters and drains at 20 liters per minute, so $ V(t) = 500 - 20t $. State a sensible domain and say what limits it.
Show the full solution

Time cannot be negative, and the tank is empty when $ 500 - 20t = 0 $, at $ t = 25 $. After that the formula gives negative volume, which is meaningless. $ 0 \le t \le 25 $ minutes
A student writes $ f(-3) = -9 $ for $ f(x) = x^2 $. Find the error.
Show the full solution

They computed $ -3^2 $, which squares 3 first and then negates. Substituting requires brackets: $ (-3)^2 = 9 $. A squared real number is never negative, which is a quick check. $ f(-3) = 9 $
A student is asked to solve $ f(x) = 8 $ for $ f(x) = 3x - 1 $ and answers 23. What did they do, and what is the correct answer?
Show the full solution

They computed $ f(8) = 3(8) - 1 = 23 $; they evaluated instead of solving. Setting $ 3x - 1 = 8 $ gives $ 3x = 9 $, so $ x = 3 $. $ x = 3 $

Topic 2.2 · Unit 2 · CA CCSS F-IF.6

Slope and rate of change

Slope is one idea wearing several names, gradient, rate of change, "per". Whenever a question says per, it is telling you a slope. The mechanical part is easy; the marks are in keeping the subtraction order consistent and interpreting the number with units.

The method

From two points: $ m = \frac{y_2 - y_1}{x_2 - x_1} $. Whichever point you call "first", use it first in both the top and the bottom. Swapping order in one only flips the sign.
From a graph: pick two points where the line crosses grid intersections exactly, then count rise over run. Guessing at a point between gridlines is where accuracy is lost.
From a table: check the $ x $-values are equally spaced before dividing differences. If they are not, use the slope formula on two rows instead.
In context: state it as "output units per input unit", dollars per hour, liters per minute. A bare number earns less than the same number with units.
Special lines: horizontal has slope 0; vertical has undefined slope. Write which one, "no slope" is ambiguous and is marked wrong.

Where marks are lost: mixing the order. For $ (2, 5) $ and $ (6, 13) $, $ \frac{13 - 5}{6 - 2} = 2 $ and $ \frac{5 - 13}{2 - 6} = 2 $ both work, but $ \frac{13 - 5}{2 - 6} = -2 $ does not. Consistency, not which point comes first, is what matters.

Worked examples

Example 1: two points with negatives. Find the slope through $ (-3, 7) $ and $ (5, -1) $.

$ m = \frac{-1 - 7}{5 - (-3)} = \frac{-8}{8} = -1 $. The $ 5 - (-3) = 8 $ step is where signs go wrong; subtracting a negative adds. A negative slope means the line falls left to right, which matches $ y $ dropping from 7 to $ -1 $.

Example 2: a table with unequal spacing. Find the rate of change from $ (1, 10), (3, 16), (7, 28) $.

The $ x $-values jump by 2 then 4, so you cannot just subtract consecutive $ y $-values. Use the formula: $ \frac{16 - 10}{3 - 1} = 3 $ and $ \frac{28 - 16}{7 - 3} = 3 $. The rate is constant at 3, so the relationship is linear after all, but only dividing by the correct run reveals it.

Example 3: interpreting in context. A phone plan's cost against minutes passes through $ (100, 32) $ and $ (250, 47) $. Interpret the slope.

$ m = \frac{47 - 32}{250 - 100} = \frac{15}{150} = 0.1 $. The units are dollars per minute, so each additional minute costs 10 cents. Saying "the slope is 0.1" without units does not answer an interpretation question.

Practice · 15 problems

1–6 are direct, 7–11 involve tables and special cases, 12–15 interpret or diagnose.

Find the slope through $ (1, 2) $ and $ (4, 11) $.
Show the full solution

$ \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3 $. $ m = 3 $
Find the slope through $ (0, 5) $ and $ (3, 5) $.
Show the full solution

$ \frac{5 - 5}{3 - 0} = 0 $. The $ y $-values never change, so the line is horizontal. $ m = 0 $
Find the slope through $ (-2, 3) $ and $ (4, -9) $.
Show the full solution

$ \frac{-9 - 3}{4 - (-2)} = \frac{-12}{6} = -2 $. $ m = -2 $
Find the slope through $ (2, 1) $ and $ (2, 8) $.
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$ \frac{8 - 1}{2 - 2} = \frac{7}{0} $, which is undefined. The line is vertical. Write "undefined", not "no slope". Undefined
A line rises 6 units for every 4 units right. What is its slope?
Show the full solution

$ \frac{6}{4} = \frac{3}{2} $. $ m = \frac{3}{2} $
Find the slope through $ (-5, -1) $ and $ (-1, -9) $.
Show the full solution

$ \frac{-9 - (-1)}{-1 - (-5)} = \frac{-8}{4} = -2 $. Both subtractions involve a negative; take them one at a time. $ m = -2 $
A table gives $ (0, 7), (2, 13), (4, 19) $. Find the rate of change and say whether it is linear.
Show the full solution

$ x $ rises by 2 each time and $ y $ by 6, so the rate is $ 6 \div 2 = 3 $ throughout. Constant rate means linear. $ m = 3 $, linear
A table gives $ (1, 4), (3, 9), (6, 18) $. Is the relationship linear?
Show the full solution

$ \frac{9-4}{3-1} = 2.5 $ but $ \frac{18-9}{6-3} = 3 $. The rate changes, so it is not linear. Subtracting $ y $-values alone would have missed this because the $ x $-spacing is unequal. Not linear
A line passes through $ (0, -4) $ with slope $ \frac{2}{3} $. Find another point on it.
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From $ (0, -4) $, move 3 right and 2 up to reach $ (3, -2) $. Any multiple works, such as $ (6, 0) $. $ (3, -2) $
Between $ t = 2 $ and $ t = 6 $ a car's distance goes from 90 km to 310 km. Find the average rate of change with units.
Show the full solution

$ \frac{310 - 90}{6 - 2} = \frac{220}{4} = 55 $. 55 km per hour
A line has slope $ -\frac{3}{4} $ and passes through $ (8, 1) $. Find the point 4 units to the right.
Show the full solution

Run of 4 means a rise of $ -\frac{3}{4} \times 4 = -3 $, so $ y $ drops from 1 to $ -2 $. $ (12, -2) $
A pool drains from 8000 liters to 5000 liters in 20 minutes. Interpret the rate of change with units, and explain the sign.
Show the full solution

$ \frac{5000 - 8000}{20} = -150 $. The pool loses 150 liters per minute; negative because the volume is decreasing. −150 liters per minute
A gym membership costs $35 to join plus $22 a month. Which number is the slope, and what does the other represent?
Show the full solution

22 is the slope, dollars per month, the rate attached to time. 35 is the $ y $-intercept, the cost at zero months. Slope 22 $/month; intercept $35
A student finds the slope through $ (3, 8) $ and $ (7, 20) $ as $ \frac{20 - 8}{3 - 7} = -3 $. Find the error.
Show the full solution

They used the second point's $ y $ on top but the first point's $ x $ first on the bottom. Keeping the order consistent gives $ \frac{20-8}{7-3} = 3 $. The line rises, so a negative answer was implausible. $ m = 3 $
Two lines have slopes 4 and $ \frac{1}{4} $. A student says they are perpendicular. Are they? Explain.
Show the full solution

No. Perpendicular slopes are negative reciprocals, the reciprocal and the sign must change. The perpendicular to 4 is $ -\frac{1}{4} $. Two positive slopes both rise, so they can never meet at a right angle. Not perpendicular

Topic 2.3 · Unit 2 · CA CCSS F-IF.7a, A-CED.2

Graphing lines and writing their equations

Three forms describe the same lines, and each exists because it makes one task easy. Choosing the right one turns most of these questions into a single substitution.

The method

Slope-intercept, $ y = mx + b $: use it to graph quickly, or when you are told the $ y $-intercept. Plot $ b $ on the $ y $-axis, then count the slope.
Point-slope, $ y - y_1 = m(x - x_1) $: use it whenever you have a slope and any point. This is the fastest route for most "write the equation" questions, and you can rearrange to $ y = mx + b $ afterwards if asked.
Standard, $ Ax + By = C $: use it to find both intercepts fast: set $ x = 0 $ for the $ y $-intercept, $ y = 0 $ for the $ x $-intercept.
From two points: find the slope first, then feed either point into point-slope. Do not try to do both steps at once.
Parallel lines share a slope. Perpendicular slopes are negative reciprocals: flip the fraction and change the sign.

Where marks are lost: point-slope with negative coordinates. Through $ (-2, 5) $ with slope 3 the equation is $ y - 5 = 3(x - (-2)) $, that is $ y - 5 = 3(x + 2) $. The minus in the formula and the minus in the coordinate combine to a plus.

Worked examples

Example 1: from two points. Write the equation of the line through $ (2, 3) $ and $ (6, 11) $ in slope-intercept form.

Slope first: $ m = \frac{11 - 3}{6 - 2} = 2 $. Now point-slope with $ (2,3) $: $ y - 3 = 2(x - 2) $. Expanding: $ y - 3 = 2x - 4 $, so $ y = 2x - 1 $.

Check with the other point: $ 2(6) - 1 = 11 $ ✓. Using the second point in point-slope gives the same final answer, which is a useful reassurance.

Example 2: perpendicular through a point. Find the line perpendicular to $ y = -\frac{2}{3}x + 4 $ passing through $ (6, -1) $.

The negative reciprocal of $ -\frac{2}{3} $ is $ \frac{3}{2} $, flip and change sign. Point-slope: $ y + 1 = \frac{3}{2}(x - 6) $. Expanding: $ y + 1 = \frac{3}{2}x - 9 $, so $ y = \frac{3}{2}x - 10 $.

Note $ y - (-1) $ became $ y + 1 $, the same sign trap as in the method note.

Example 3: standard form and intercepts. Find both intercepts of $ 3x + 4y = 24 $ and sketch it.

Set $ x = 0 $: $ 4y = 24 $, so $ y = 6 $, the point $ (0,6) $. Set $ y = 0 $: $ 3x = 24 $, so $ x = 8 $, the point $ (8,0) $. Plot those two and join them. This is faster than converting to slope-intercept when a sketch is all that is wanted.

Practice · 15 problems

1–6 are direct, 7–12 combine forms and conditions, 13–15 are context and diagnosis.

State the slope and $ y $-intercept of $ y = -3x + 7 $.
Show the full solution

Read them straight off. $ m = -3 $, intercept $ (0, 7) $
Write the equation of the line with slope 5 and $ y $-intercept $ -2 $.
Show the full solution

$ y = 5x - 2 $
Write, in point-slope form, the line through $ (4, 9) $ with slope 2.
Show the full solution

$ y - 9 = 2(x - 4) $
Write, in point-slope form, the line through $ (-3, 6) $ with slope $ -1 $.
Show the full solution

$ x - (-3) $ becomes $ x + 3 $. $ y - 6 = -(x + 3) $
Find both intercepts of $ 2x + 5y = 20 $.
Show the full solution

$ x = 0 $ gives $ y = 4 $; $ y = 0 $ gives $ x = 10 $. $ (0,4) $ and $ (10,0) $
Convert $ 4x - 2y = 8 $ to slope-intercept form.
Show the full solution

$ -2y = -4x + 8 $, then divide by $ -2 $: $ y = 2x - 4 $. Dividing by a negative changes both signs. $ y = 2x - 4 $
Write the equation of the line through $ (1, 5) $ and $ (3, 11) $ in slope-intercept form.
Show the full solution

$ m = \frac{11-5}{3-1} = 3 $. Then $ y - 5 = 3(x - 1) $, so $ y = 3x + 2 $. Check: $ 3(3) + 2 = 11 $ ✓. $ y = 3x + 2 $
Write the equation of the line parallel to $ y = 4x - 1 $ through $ (2, 3) $.
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Parallel means the same slope, 4. $ y - 3 = 4(x - 2) $, so $ y = 4x - 5 $. $ y = 4x - 5 $
Write the equation of the line perpendicular to $ y = 2x + 3 $ through $ (4, 1) $.
Show the full solution

Negative reciprocal of 2 is $ -\frac{1}{2} $. $ y - 1 = -\frac{1}{2}(x - 4) $, so $ y = -\frac{1}{2}x + 3 $. $ y = -\frac{1}{2}x + 3 $
Write the equation of the horizontal line through $ (5, -2) $.
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Horizontal means $ y $ never changes. $ y = -2 $
Write the equation of the vertical line through $ (5, -2) $.
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Vertical means $ x $ never changes. It cannot be written as $ y = mx + b $ because its slope is undefined. $ x = 5 $
Write the equation of the line through $ (-2, 5) $ and $ (2, -3) $.
Show the full solution

$ m = \frac{-3 - 5}{2 - (-2)} = \frac{-8}{4} = -2 $. Point-slope with $ (2,-3) $: $ y + 3 = -2(x - 2) $, so $ y = -2x + 1 $. Check with $ (-2,5) $: $ -2(-2) + 1 = 5 $ ✓. $ y = -2x + 1 $
A taxi charges $3.50 plus $2 a mile. Write the cost as a linear equation and state what each part means.
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$ C = 2m + 3.50 $. The slope 2 is dollars per mile; the intercept 3.50 is the charge before traveling any distance. $ C = 2m + 3.50 $
A candle is 30 cm tall and burns 2 cm per hour. Write an equation for its height and find when it is fully burned.
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$ h = 30 - 2t $. It reaches zero when $ 30 - 2t = 0 $, at $ t = 15 $. That is the $ x $-intercept of the line. $ h = 30 - 2t $; 15 hours
A student writes the line through $ (-4, 2) $ with slope 3 as $ y - 2 = 3(x - 4) $. Find the error.
Show the full solution

They used $ +4 $ instead of $ -4 $ for $ x_1 $. Point-slope subtracts the coordinate, so $ x - (-4) = x + 4 $. Correct: $ y - 2 = 3(x + 4) $, which expands to $ y = 3x + 14 $. Testing $ x = -4 $ in their version gives $ y = -22 $, not 2, which exposes it immediately. $ y - 2 = 3(x + 4) $

Topic 2.4 · Unit 2 · CA CCSS F-IF.7b

Piecewise, step and absolute value functions

A piecewise function is several rules sharing one name, each valid on its own stretch of the $ x $-axis. Everything here follows from one habit: decide which piece applies before you calculate anything.

The method

To evaluate: read the conditions first and find the interval your input falls in. Only then substitute, into that rule alone. Substituting into every piece and picking a favorite is the commonest error.
At a boundary: the condition tells you which piece owns it. If one piece says $ x < 4 $ and the next says $ x \ge 4 $, then $ x = 4 $ belongs to the second.
To graph: graph each rule but draw it only across its own interval. Put a closed circle where the endpoint is included ($ \le, \ge $) and an open circle where it is not ($ <, > $).
Absolute value $ y = |x| $ is a V with its vertex at the origin. For $ y = a|x - h| + k $ the vertex moves to $ (h, k) $; a negative $ a $ flips the V upside down.
Step functions jump rather than slope, parking charges, postage rates. Their graphs are flat segments with open and closed ends.

Where marks are lost: the vertex shift inside the bars. $ y = |x - 3| $ moves the V right 3, not left, the same counter-intuitive sign as every other horizontal shift. Test it: at $ x = 3 $ the expression is zero, which is the lowest the absolute value can be, so the vertex must sit there.

Worked examples

Example 1: evaluating including a boundary. For

$ f(x) = \begin{cases} 2x + 1 & x < 3 \\ 10 - x & x \ge 3 \end{cases} $

find $ f(0) $, $ f(3) $ and $ f(7) $.

$ f(0) $: 0 is less than 3, so use the first rule, $ 2(0)+1 = 1 $. $ f(3) $: the condition $ x \ge 3 $ includes 3, so use the second, $ 10 - 3 = 7 $. $ f(7) $: also the second, $ 10 - 7 = 3 $. Using $ 2(3)+1 = 7 $ for $ f(3) $ happens to agree here, but that is a coincidence of this function, not a method.

Example 2: absolute value with a shift. Describe the graph of $ y = -2|x + 1| + 5 $.

Rewrite the inside as $ x - (-1) $, so $ h = -1 $ and the vertex is at $ (-1, 5) $. The $ -2 $ makes the V open downward and twice as steep. So it is an upside-down V peaking at $ (-1, 5) $, falling with slope $ -2 $ to the right and $ +2 $ to the left.

Example 3: writing a piecewise rule. A car park charges $4 for the first hour or part of it, then $2 for each additional hour or part. Write a rule for the first three hours.

Any time up to and including 1 hour costs 4; above 1 up to 2 costs 6; above 2 up to 3 costs 8:

$ C(t) = \begin{cases} 4 & 0 < t \le 1 \\ 6 & 1 < t \le 2 \\ 8 & 2 < t \le 3 \end{cases} $

"Or part of it" is what makes this a step function, the charge jumps the instant you pass a boundary rather than rising gradually.

Practice · 15 problems

Use $ f(x) = \begin{cases} x + 4 & x < 2 \\ 3x - 2 & x \ge 2 \end{cases} $ for problems 1–4.

Find $ f(0) $.
Show the full solution

0 is less than 2, so the first rule: $ 0 + 4 = 4 $. $ f(0) = 4 $
Find $ f(2) $.
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The condition $ x \ge 2 $ includes 2, so the second rule: $ 3(2) - 2 = 4 $. $ f(2) = 4 $
Find $ f(-3) $.
Show the full solution

First rule: $ -3 + 4 = 1 $. $ f(-3) = 1 $
Find $ f(5) $.
Show the full solution

Second rule: $ 3(5) - 2 = 13 $. $ f(5) = 13 $
Evaluate $ |{-7}| $ and $ -|7| $.
Show the full solution

$ |{-7}| = 7 $, the bars strip the sign. $ -|7| = -7 $, the negative is outside, so it applies afterwards. 7 and $ -7 $
State the vertex of $ y = |x - 5| $.
Show the full solution

The inside is zero at $ x = 5 $, which is the minimum. $ (5, 0) $
State the vertex of $ y = |x + 2| - 3 $.
Show the full solution

$ x + 2 = x - (-2) $, so $ h = -2 $, and $ k = -3 $. $ (-2, -3) $
Describe how $ y = -|x| $ differs from $ y = |x| $.
Show the full solution

The negative reflects it across the $ x $-axis, so the V opens downward with its maximum at the origin instead of its minimum. Reflected; opens downward
Solve $ |x| = 6 $.
Show the full solution

Two numbers are 6 units from zero. $ x = 6 $ or $ x = -6 $
Solve $ |x - 3| = 5 $.
Show the full solution

The inside is 5 or $ -5 $: $ x - 3 = 5 $ gives $ x = 8 $; $ x - 3 = -5 $ gives $ x = -2 $. $ x = 8 $ or $ x = -2 $
For $ g(x) = \begin{cases} -x & x \le 0 \\ x^2 & x > 0 \end{cases} $, find $ g(-4) $ and $ g(4) $.
Show the full solution

$ -4 \le 0 $, so $ g(-4) = -(-4) = 4 $. $ 4 > 0 $, so $ g(4) = 16 $. 4 and 16
Describe the graph of $ y = 3|x - 1| + 2 $.
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Vertex at $ (1, 2) $, opening upward, three times steeper than $ y = |x| $, slope 3 to the right and $ -3 $ to the left. V with vertex $ (1,2) $, steepness 3
A courier charges $5 for parcels up to 1 kg, $8 up to 3 kg, and $12 up to 5 kg. Write a piecewise rule.
Show the full solution

$ C(w) = \begin{cases} 5 & 0 < w \le 1 \\ 8 & 1 < w \le 3 \\ 12 & 3 < w \le 5 \end{cases} $. "Up to" means the boundary is included, so each interval closes at its right end. As above
Using the courier rule, what does a 3 kg parcel cost, and what does a 3.1 kg parcel cost?
Show the full solution

3 kg falls in $ 1 < w \le 3 $, so $8. 3.1 kg crosses into the next band, so $12. A tenth of a kilogram adds $4; that jump is exactly what makes it a step function. $8 and $12
A student says the vertex of $ y = |x + 6| $ is $ (6, 0) $. Find the error.
Show the full solution

They read the sign straight off instead of matching the form $ |x - h| $. Here $ x + 6 = x - (-6) $, so $ h = -6 $. Check: at $ x = -6 $ the expression is 0, its smallest value; at $ x = 6 $ it is 12. Vertex $ (-6, 0) $

Unit 2 mixed review · 10 problems · all topics

Unit 2 mixed review: Linear Functions

Function notation, slope, the three forms of a line, and piecewise functions, shuffled.

For $ f(x) = 5 - 2x $, find $ f(-3) $.
Show the full solution

$ 5 - 2(-3) = 5 + 6 = 11 $. 11
Find the slope through $ (-1, 4) $ and $ (3, -8) $.
Show the full solution

$ \frac{-8 - 4}{3 - (-1)} = \frac{-12}{4} = -3 $. $ m = -3 $
Write the equation of the line through $ (2, 7) $ and $ (5, 16) $.
Show the full solution

$ m = \frac{16-7}{5-2} = 3 $. Point-slope: $ y - 7 = 3(x - 2) $, so $ y = 3x + 1 $. Check: $ 3(5)+1 = 16 $ ✓. $ y = 3x + 1 $
For $ f(x) = 3x - 4 $, solve $ f(x) = 11 $.
Show the full solution

$ 3x - 4 = 11 $, so $ x = 5 $. $ x = 5 $
Give the vertex of $ y = |x - 2| + 7 $.
Show the full solution

Right 2, up 7. $ (2, 7) $
Write the line perpendicular to $ y = \frac{1}{3}x - 2 $ through $ (3, 5) $.
Show the full solution

Negative reciprocal of $ \frac{1}{3} $ is $ -3 $. $ y - 5 = -3(x - 3) $, so $ y = -3x + 14 $. $ y = -3x + 14 $
Find both intercepts of $ 5x - 2y = 20 $.
Show the full solution

$ x = 0 $: $ y = -10 $. $ y = 0 $: $ x = 4 $. $ (0,-10) $ and $ (4,0) $
For $ f(x) = \begin{cases} 2x & x < 1 \\ x + 5 & x \ge 1 \end{cases} $, find $ f(0) $ and $ f(1) $.
Show the full solution

$ 0 < 1 $ uses the first rule: 0. $ x = 1 $ satisfies $ x \ge 1 $, so the second: $ 6 $. 0 and 6
A pool drains from 12 000 to 9000 liters in 25 minutes. Interpret the rate of change.
Show the full solution

$ \frac{9000 - 12000}{25} = -120 $. It loses 120 liters per minute; negative because the volume falls. −120 liters per minute
Solve $ |x - 4| = 9 $.
Show the full solution

$ x - 4 = 9 $ gives 13; $ x - 4 = -9 $ gives $ -5 $. $ x = 13 $ or $ x = -5 $

Topic 3.1 · Unit 3 · CA CCSS A-REI.6, A-REI.11

Solving systems by graphing

A solution to a system is a point that satisfies both equations at once, which on a graph means a point lying on both lines, their intersection. Graphing is the method that shows you why a system can have no solution or infinitely many, which is why it is taught first even though it is the least precise.

The method

Put both equations in $ y = mx + b $ form unless they are already easy to plot.
Graph both on the same axes, plot the intercept, then count the slope.
Read the intersection and write it as an ordered pair $ (x, y) $, not as a single number.
Check in both equations. A point that satisfies only one is not a solution, and graphing is imprecise enough that this check is not optional.
Recognize the three outcomes before you graph, from the slopes and intercepts alone:
- Different slopes → the lines cross once → one solution
- Same slope, different intercept → parallel, never meet → no solution
- Same slope, same intercept → the same line → infinitely many

Where marks are lost: answering with only the $ x $-value. A system in two variables has a solution that is a point, so both coordinates are required. Also, graphing only gives exact answers when the intersection lands on a grid point, if it looks like $ (2.3, 4.7) $, solve algebraically instead.

Worked examples

Example 1: one solution. Solve by graphing: $ y = 2x - 1 $ and $ y = -x + 5 $.

The first has intercept $ (0,-1) $ and rises 2 for every 1 right. The second has intercept $ (0,5) $ and falls 1 for every 1 right. They cross at $ (2, 3) $.

Check both: $ 2(2) - 1 = 3 $ ✓ and $ -(2) + 5 = 3 $ ✓. The slopes differ, so exactly one intersection was guaranteed before any plotting.

Example 2: no solution. Solve: $ y = 3x + 2 $ and $ 6x - 2y = 10 $.

Rearrange the second: $ -2y = -6x + 10 $, so $ y = 3x - 5 $. Both have slope 3 but different intercepts, so the lines are parallel and never meet. No solution. Recognizing this from the rearranged form takes seconds and saves drawing anything.

Example 3: infinitely many. Solve: $ y = \frac{1}{2}x + 3 $ and $ 2y - x = 6 $.

Rearrange the second: $ 2y = x + 6 $, so $ y = \frac{1}{2}x + 3 $. Identical to the first, the two equations describe the same line, so every point on it satisfies both. Infinitely many solutions. Writing "all real numbers" is not quite right; the solutions are all the points on that line.

Practice · 15 problems

1–6 are direct, 7–11 require rearranging first, 12–15 involve context and diagnosis.

Is $ (3, 5) $ a solution of $ y = x + 2 $ and $ y = 2x - 1 $?
Show the full solution

$ 3 + 2 = 5 $ ✓ and $ 2(3) - 1 = 5 $ ✓. It satisfies both. Yes
Is $ (1, 4) $ a solution of $ y = 3x + 1 $ and $ y = x + 2 $?
Show the full solution

$ 3(1) + 1 = 4 $ ✓ but $ 1 + 2 = 3 \neq 4 $ ✗. It must satisfy both. No
Solve by graphing: $ y = x + 1 $ and $ y = -x + 5 $.
Show the full solution

Setting them equal: $ x + 1 = -x + 5 $, so $ 2x = 4 $ and $ x = 2 $, giving $ y = 3 $. $ (2, 3) $
Solve by graphing: $ y = 2x $ and $ y = x + 3 $.
Show the full solution

$ 2x = x + 3 $ gives $ x = 3 $, so $ y = 6 $. $ (3, 6) $
How many solutions does $ y = 4x + 1 $ and $ y = 4x - 3 $ have?
Show the full solution

Same slope 4, different intercepts, parallel lines. No solution
How many solutions does $ y = -2x + 6 $ and $ y = -2x + 6 $ have?
Show the full solution

Identical equations, so the same line. Infinitely many
Without graphing, decide how many solutions $ y = 5x - 2 $ and $ 10x - 2y = 4 $ have.
Show the full solution

Rearrange: $ -2y = -10x + 4 $, so $ y = 5x - 2 $. Same line. Infinitely many
Without graphing, decide how many solutions $ 3x + y = 7 $ and $ 6x + 2y = 9 $ have.
Show the full solution

First: $ y = -3x + 7 $. Second: $ 2y = -6x + 9 $, so $ y = -3x + 4.5 $. Same slope, different intercepts. No solution
Solve: $ x + y = 6 $ and $ y = 2x $.
Show the full solution

Substituting the second into the first: $ x + 2x = 6 $, so $ x = 2 $ and $ y = 4 $. $ (2, 4) $
Solve: $ 2x + y = 8 $ and $ y = x - 1 $.
Show the full solution

$ 2x + (x - 1) = 8 $, so $ 3x = 9 $, $ x = 3 $ and $ y = 2 $. $ (3, 2) $
Two lines have slopes 3 and $ -3 $. What can you say about the number of solutions?
Show the full solution

Different slopes, so they must cross exactly once whatever the intercepts are. Exactly one solution
A gym charges $30 joining plus $20 a month; a rival charges $10 a month with no joining fee. After how many months is the cost the same?
Show the full solution

$ 30 + 20m = 10m $ gives $ 10m = -30 $, so $ m = -3 $. A negative month is impossible, meaning the costs are never equal going forward, the second gym is always cheaper. Never; the rival is always cheaper
Phone plan A costs $25 plus $0.10 per minute; plan B costs $15 plus $0.20 per minute. At how many minutes are they equal?
Show the full solution

$ 25 + 0.1t = 15 + 0.2t $, so $ 10 = 0.1t $ and $ t = 100 $. At 100 minutes both cost $35; below that B is cheaper, above it A is. 100 minutes
A student solves a system by graphing and answers "$ x = 4 $". Why is that incomplete?
Show the full solution

A system in two variables has a point as its solution, so the $ y $-coordinate is needed too. The answer must be an ordered pair such as $ (4, 7) $. Needs both coordinates
A student graphs $ y = 2x + 1 $ and $ y = 2x + 1 $, sees one line, and writes "no solution". Find the error.
Show the full solution

They confused the two special cases. Seeing one line means the equations are the same line, so every point on it works, infinitely many solutions. No solution is what two parallel lines give. Infinitely many

Topic 3.2 · Unit 3 · CA CCSS A-REI.5–6

Solving systems by substitution and elimination

Both methods do the same thing: get rid of one variable so you are left with an ordinary one-variable equation. Which one is faster is decided by how the system is written, and recognizing that in two seconds saves most of the work.

The method

Choosing: if one equation already has a variable alone ($ y = \dots $ or $ x = \dots $) or has a coefficient of 1, substitute. If the variables are lined up in columns with awkward coefficients, eliminate.

Substitution: isolate one variable in one equation; substitute that whole expression into the other equation, in brackets; solve; then back-substitute to find the second variable.
Elimination: line the equations up; multiply one or both so that one variable has coefficients that are opposites; add the equations to cancel it; solve; then back-substitute.
Back-substitution is not optional. Finding $ x $ is half the answer. Put it into whichever original equation is simpler.
Check in both originals. An arithmetic slip in the middle is invisible until you do.
Same special cases as graphing: if both variables vanish and leave something true, infinitely many; something false, no solution.

Where marks are lost: substituting into the same equation you rearranged, which collapses to $ 0 = 0 $ and tells you nothing. And forgetting brackets: substituting $ y = x - 3 $ into $ 2x - y = 8 $ gives $ 2x - (x - 3) $, which is $ 2x - x + 3 $, not $ 2x - x - 3 $.

Worked examples

Example 1: substitution with the bracket trap. Solve $ y = x - 3 $ and $ 2x - y = 8 $.

The first is already isolated, so substitute into the second, in brackets: $ 2x - (x - 3) = 8 $. Distribute the minus: $ 2x - x + 3 = 8 $, so $ x + 3 = 8 $ and $ x = 5 $. Back-substitute: $ y = 5 - 3 = 2 $.

Check: $ 2(5) - 2 = 8 $ ✓. $ (5, 2) $

Example 2: elimination needing one multiplication. Solve $ 3x + 2y = 16 $ and $ x - 2y = 4 $.

The $ y $ coefficients are already opposites, $ +2 $ and $ -2 $, so just add: $ 4x = 20 $, giving $ x = 5 $. Back-substitute into the second: $ 5 - 2y = 4 $, so $ -2y = -1 $ and $ y = 0.5 $.

Check in the first: $ 3(5) + 2(0.5) = 16 $ ✓. $ (5, 0.5) $

Example 3: elimination needing both multiplied. Solve $ 2x + 3y = 12 $ and $ 5x - 2y = 11 $.

Nothing matches. To eliminate $ y $, make the coefficients 6 and $ -6 $: multiply the first by 2 and the second by 3.

$ 4x + 6y = 24 $ and $ 15x - 6y = 33 $. Adding: $ 19x = 57 $, so $ x = 3 $. Back-substitute: $ 2(3) + 3y = 12 $, so $ 3y = 6 $ and $ y = 2 $.

Check in the second: $ 5(3) - 2(2) = 11 $ ✓. $ (3, 2) $

Practice · 15 problems

1–5 substitution, 6–10 elimination, 11–15 your choice of method plus word problems.

Solve: $ y = 2x $ and $ x + y = 9 $.
Show the full solution

$ x + 2x = 9 $, so $ x = 3 $ and $ y = 6 $. $ (3, 6) $
Solve: $ x = y + 4 $ and $ 2x + y = 14 $.
Show the full solution

$ 2(y + 4) + y = 14 $, so $ 3y + 8 = 14 $, $ y = 2 $ and $ x = 6 $. $ (6, 2) $
Solve: $ y = 3x - 1 $ and $ y = x + 7 $.
Show the full solution

Both are already $ y = \dots $, so set them equal: $ 3x - 1 = x + 7 $, giving $ x = 4 $ and $ y = 11 $. $ (4, 11) $
Solve: $ y = x - 5 $ and $ 3x - y = 11 $.
Show the full solution

$ 3x - (x - 5) = 11 $, so $ 3x - x + 5 = 11 $, $ 2x = 6 $, $ x = 3 $ and $ y = -2 $. $ (3, -2) $
Solve: $ x + 2y = 11 $ and $ x = 3y + 1 $.
Show the full solution

$ (3y + 1) + 2y = 11 $, so $ 5y = 10 $, $ y = 2 $ and $ x = 7 $. $ (7, 2) $
Solve: $ x + y = 10 $ and $ x - y = 4 $.
Show the full solution

Adding eliminates $ y $: $ 2x = 14 $, so $ x = 7 $ and $ y = 3 $. $ (7, 3) $
Solve: $ 2x + 3y = 13 $ and $ 2x - y = 1 $.
Show the full solution

Subtracting eliminates $ x $: $ 4y = 12 $, so $ y = 3 $, then $ 2x - 3 = 1 $ gives $ x = 2 $. $ (2, 3) $
Solve: $ 3x + y = 14 $ and $ 2x - 3y = -9 $.
Show the full solution

Multiply the first by 3: $ 9x + 3y = 42 $. Adding to the second: $ 11x = 33 $, so $ x = 3 $ and $ y = 14 - 9 = 5 $. $ (3, 5) $
Solve: $ 3x + 4y = 10 $ and $ 2x - 3y = 1 $.
Show the full solution

Nothing matches, so both need multiplying. To eliminate $ y $, aim for 12 and $ -12 $: multiply the first by 3 and the second by 4.

$ 9x + 12y = 30 $ and $ 8x - 12y = 4 $. Adding: $ 17x = 34 $, so $ x = 2 $. Back-substitute into the first original: $ 6 + 4y = 10 $, giving $ y = 1 $.

Check in the second: $ 2(2) - 3(1) = 1 $ ✓. $ (2, 1) $
Solve: $ 3x - 2y = 8 $ and $ 5x + 2y = 24 $.
Show the full solution

The $ y $ terms are already opposites, so add straight away: $ 8x = 32 $, giving $ x = 4 $. Then $ 3(4) - 2y = 8 $, so $ -2y = -4 $ and $ y = 2 $.

Check in the second: $ 5(4) + 2(2) = 24 $ ✓. $ (4, 2) $
Solve by whichever method is faster: $ y = 4 - x $ and $ 3x + 2y = 11 $.
Show the full solution

Substitution, since $ y $ is already isolated. $ 3x + 2(4 - x) = 11 $, so $ 3x + 8 - 2x = 11 $, $ x = 3 $ and $ y = 1 $. $ (3, 1) $
Two numbers add to 42 and differ by 8. Find them.
Show the full solution

$ x + y = 42 $, $ x - y = 8 $. Adding: $ 2x = 50 $, so $ x = 25 $ and $ y = 17 $. 25 and 17
Adult tickets cost $12 and child tickets $7. 200 tickets sold for $1900 in total. How many of each?
Show the full solution

$ a + c = 200 $ and $ 12a + 7c = 1900 $. From the first, $ c = 200 - a $: $ 12a + 7(200 - a) = 1900 $, so $ 5a + 1400 = 1900 $, $ a = 100 $ and $ c = 100 $. 100 adult, 100 child
Solve: $ 2x + 4y = 10 $ and $ x + 2y = 5 $.
Show the full solution

The first is exactly twice the second, so they are the same line. Eliminating gives $ 0 = 0 $, which is true. Infinitely many solutions
A student substitutes $ y = x - 3 $ into $ 2x - y = 8 $ and writes $ 2x - x - 3 = 8 $. Find the error.
Show the full solution

They dropped the brackets. The whole expression is being subtracted: $ 2x - (x - 3) = 2x - x + 3 $. Their version gives $ x = 11 $; the correct working gives $ x = 5 $. Checking $ (11, 8) $ in the original fails, which would have caught it. $ x = 5 $, so $ (5, 2) $

Topic 3.3 · Unit 3 · CA CCSS A-REI.12

Systems of linear inequalities

One linear inequality shades half the plane. A system shades the region where every inequality is satisfied at once, the overlap. The algebra is the same as Unit 1; the new skills are choosing the boundary style and shading the correct side.

The method

Rearrange each to $ y = $ form so you can see the boundary line and which way to shade.
Draw the boundary. Dashed for $ < $ or $ > $, points on the line do not count. Solid for $ \le $ or $ \ge $; they do.
Shade. Once in $ y = $ form, $ y > $ shades above the line and $ y < $ shades below.
Test a point to be sure, $ (0,0) $ is easiest whenever the boundary does not pass through it. If it satisfies the inequality, shade its side.
The solution is the overlap only. Shade each inequality lightly and take the region covered by all of them. A point in just one region is not a solution.

Where marks are lost: forgetting to flip when rearranging. $ -2y > 4x - 6 $ becomes $ y < -2x + 3 $, dividing by $ -2 $ reverses the sign, which also reverses which side you shade. Testing $ (0,0) $ catches this instantly.

Worked examples

Example 1: boundary style and shading. Graph $ y \ge 2x - 3 $.

The boundary is $ y = 2x - 3 $, drawn solid because $ \ge $ includes it. Test $ (0,0) $: is $ 0 \ge 2(0) - 3 $, that is $ 0 \ge -3 $? Yes, so shade the side containing the origin, above the line, which matches $ y \ge $.

Example 2: a rearrangement that flips. Graph $ -3y > 6x - 9 $.

Divide by $ -3 $ and reverse: $ y < -2x + 3 $. The boundary $ y = -2x + 3 $ is dashed. Test $ (0,0) $: is $ 0 < 3 $? Yes, so shade below, toward the origin. Had you divided without flipping you would have shaded the wrong half.

Example 3: a system. Graph $ y < x + 2 $ and $ y \ge -x $.

First: dashed line $ y = x + 2 $, shade below. Second: solid line $ y = -x $, shade above. The solution is the wedge where both shadings overlap, below the dashed line and above the solid one, meeting at $ (-1, 1) $.

Test $ (2, 1) $: $ 1 < 4 $ ✓ and $ 1 \ge -2 $ ✓, so it is in the region. Test $ (0, 5) $: $ 5 < 2 $ ✗, so it is not, despite satisfying the second.

Practice · 15 problems

1–6 are single inequalities, 7–11 are systems, 12–15 involve context and diagnosis.

Should the boundary of $ y > 3x + 1 $ be solid or dashed?
Show the full solution

$ > $ excludes the line itself. Dashed
Should the boundary of $ y \le -x + 4 $ be solid or dashed?
Show the full solution

$ \le $ includes the line. Solid
For $ y < 2x $, do you shade above or below the boundary?
Show the full solution

In $ y = $ form, $ y < $ means below. Below
Is $ (1, 5) $ a solution of $ y \ge 4x $?
Show the full solution

$ 5 \ge 4(1) = 4 $ ✓. Yes
Is $ (0, 0) $ a solution of $ y > x + 2 $?
Show the full solution

$ 0 > 2 $ is false. No
Rearrange $ 2x + y < 6 $ into $ y = $ form and say which way to shade.
Show the full solution

$ y < -2x + 6 $. Subtracting $ 2x $ does not flip anything. Shade below, dashed boundary
Rearrange $ -y \ge 3x - 5 $ and say which way to shade.
Show the full solution

Multiplying by $ -1 $ flips: $ y \le -3x + 5 $. Shade below, solid boundary
Is $ (2, 1) $ a solution of the system $ y < x + 3 $ and $ y \ge x - 4 $?
Show the full solution

$ 1 < 5 $ ✓ and $ 1 \ge -2 $ ✓. Both hold. Yes
Is $ (0, 6) $ a solution of $ y < x + 3 $ and $ y \ge x - 4 $?
Show the full solution

$ 6 < 3 $ is false, so it fails the first even though $ 6 \ge -4 $ holds. All inequalities must be satisfied. No
Describe the solution region of $ y > 1 $ and $ x \le 4 $.
Show the full solution

$ y = 1 $ is a horizontal dashed line, shaded above; $ x = 4 $ is a vertical solid line, shaded left. The overlap is the region above $ y = 1 $ and left of $ x = 4 $. Upper-left region bounded by both
Give one point in the solution region of $ y \ge 0 $, $ x \ge 0 $ and $ x + y \le 5 $.
Show the full solution

The region is a triangle with vertices $ (0,0) $, $ (5,0) $, $ (0,5) $. Any point inside works, for example $ (1, 1) $: all three hold. $ (1, 1) $
A student has at most $40 and buys notebooks at $4 and pens at $2. Write a system for what they can buy, including sensible restrictions.
Show the full solution

$ 4n + 2p \le 40 $, with $ n \ge 0 $ and $ p \ge 0 $ since you cannot buy a negative number. Whole numbers only, so the real solutions are the lattice points in the region. $ 4n + 2p \le 40, n \ge 0, p \ge 0 $
Using that system, can they buy 6 notebooks and 9 pens?
Show the full solution

$ 4(6) + 2(9) = 24 + 18 = 42 $, which exceeds 40. No, $2 short
A student graphs $ y > 2x - 1 $ with a solid line. What is wrong?
Show the full solution

A strict inequality excludes the boundary, so the line must be dashed. A solid line wrongly claims that points on the line are solutions, but there $ y = 2x - 1 $, not $ y > 2x - 1 $. Should be dashed
A student rearranges $ -2y < 6x - 4 $ to $ y < -3x + 2 $ and shades below. Find the error.
Show the full solution

Dividing by $ -2 $ must reverse the inequality: $ y > -3x + 2 $, so the shading goes above. Testing $ (0,0) $ in the original (is $ 0 < -4 $? No) shows the origin is not in the region, but their shading includes it. $ y > -3x + 2 $, shade above

Unit 3 mixed review · 10 problems · all topics

Unit 3 mixed review: Systems

Graphing, substitution, elimination and inequalities, with the method left for you to choose.

Solve: $ y = 3x $ and $ x + y = 12 $.
Show the full solution

$ x + 3x = 12 $, so $ x = 3 $ and $ y = 9 $. $ (3, 9) $
Solve: $ 2x + y = 11 $ and $ 2x - y = 5 $.
Show the full solution

Adding eliminates $ y $: $ 4x = 16 $, so $ x = 4 $ and $ y = 3 $. $ (4, 3) $
How many solutions does $ y = 2x + 5 $ and $ 4x - 2y = 3 $ have?
Show the full solution

Second rearranges to $ y = 2x - 1.5 $. Same slope, different intercept. No solution
Solve: $ 3x + 2y = 16 $ and $ x - y = 3 $.
Show the full solution

From the second, $ x = y + 3 $. Substituting: $ 3(y+3) + 2y = 16 $, so $ 5y = 7 $ and $ y = 1.4 $, giving $ x = 4.4 $. $ (4.4, 1.4) $
Is $ (3, 2) $ a solution of $ y < x $ and $ y \ge x - 4 $?
Show the full solution

$ 2 < 3 $ ✓ and $ 2 \ge -1 $ ✓. Yes
Should the boundary of $ y \ge 2x - 1 $ be solid or dashed, and which side is shaded?
Show the full solution

$ \ge $ includes the line, so solid, and $ y \ge $ shades above. Solid, shade above
Solve: $ 4x + 3y = 10 $ and $ 2x - 3y = 8 $.
Show the full solution

The $ y $ terms are opposites, so add: $ 6x = 18 $, $ x = 3 $, then $ 12 + 3y = 10 $ gives $ y = -\frac{2}{3} $. $ (3, -\frac{2}{3}) $
Two numbers add to 30; one is four times the other. Find them.
Show the full solution

$ x + y = 30 $, $ x = 4y $. So $ 5y = 30 $, $ y = 6 $ and $ x = 24 $. 24 and 6
Rearrange $ -4y > 8x - 12 $ into $ y = $ form.
Show the full solution

Dividing by $ -4 $ flips the sign: $ y < -2x + 3 $. $ y < -2x + 3 $
Tickets cost $9 for adults and $5 for children. 240 tickets raised $1660. How many of each?
Show the full solution

$ a + c = 240 $ and $ 9a + 5c = 1660 $. Substituting $ c = 240 - a $: $ 9a + 1200 - 5a = 1660 $, so $ 4a = 460 $ and $ a = 115 $, $ c = 125 $. 115 adult, 125 child

Topic 4.1 · Unit 4 · CA CCSS F-BF.2, F-LE.2

Arithmetic and geometric sequences

Arithmetic sequences add the same amount each step; geometric sequences multiply by the same amount. Identifying which you have takes one subtraction and one division, and everything after that follows a formula.

The method

Identify the type. Subtract consecutive terms: if the difference is constant it is arithmetic, and that constant is $ d $. Divide consecutive terms: if the ratio is constant it is geometric, and that constant is $ r $.
Arithmetic explicit rule: $ a_n = a_1 + (n-1)d $. The $ (n-1) $ is there because the first term has had no steps added to it yet.
Geometric explicit rule: $ a_n = a_1 \cdot r^{\,n-1} $, for the same reason, the first term has been multiplied zero times.
Recursive rules describe the step rather than the position: $ a_n = a_{n-1} + d $ or $ a_n = a_{n-1} \cdot r $. Either form must state $ a_1 $ as well, or it describes nothing.
Connect to functions. An arithmetic sequence is a linear function on whole-number inputs, with $ d $ as its slope. A geometric sequence is an exponential function, with $ r $ as its base.

Where marks are lost: the $ n-1 $. To find the 10th term of an arithmetic sequence starting at 5 with $ d = 3 $, it is $ 5 + 9(3) = 32 $, not $ 5 + 10(3) = 35 $. Count the gaps, not the terms, ten terms have nine gaps between them.

Worked examples

Example 1: arithmetic, explicit rule and a distant term. For $ 7, 11, 15, 19, \dots $ write the rule and find the 25th term.

Differences: $ 11-7 = 4 $, $ 15-11 = 4 $. Constant, so arithmetic with $ d = 4 $ and $ a_1 = 7 $. Rule: $ a_n = 7 + (n-1)4 $, which simplifies to $ a_n = 4n + 3 $.

$ a_{25} = 4(25) + 3 = 103 $. Check with the unsimplified form: $ 7 + 24(4) = 103 $ ✓.

Example 2: geometric. For $ 3, 12, 48, 192, \dots $ write the rule and find the 7th term.

Ratios: $ 12 \div 3 = 4 $, $ 48 \div 12 = 4 $. Geometric with $ r = 4 $, $ a_1 = 3 $. Rule: $ a_n = 3 \cdot 4^{\,n-1} $.

$ a_7 = 3 \cdot 4^6 = 3 \times 4096 = 12288 $. The exponent is 6, not 7, six multiplications get you from the first term to the seventh.

Example 3: working backwards. An arithmetic sequence has $ a_4 = 14 $ and $ a_9 = 34 $. Find $ a_1 $ and the rule.

From the 4th term to the 9th is five gaps, and the value rose by $ 34 - 14 = 20 $, so $ d = 20 \div 5 = 4 $. To get back from $ a_4 $ to $ a_1 $ is three gaps down: $ 14 - 3(4) = 2 $. Rule: $ a_n = 2 + (n-1)4 = 4n - 2 $.

Check: $ a_4 = 4(4) - 2 = 14 $ ✓ and $ a_9 = 4(9) - 2 = 34 $ ✓.

Practice · 15 problems

1–6 identify and continue, 7–11 build rules, 12–15 work backwards or into context.

Is $ 2, 5, 8, 11, \dots $ arithmetic or geometric? Give $ d $ or $ r $.
Show the full solution

Differences are all 3; ratios are not constant. Arithmetic, $ d = 3 $
Is $ 2, 6, 18, 54, \dots $ arithmetic or geometric? Give $ d $ or $ r $.
Show the full solution

Ratios are all 3. Geometric, $ r = 3 $
Write the next two terms of $ 20, 17, 14, \dots $
Show the full solution

$ d = -3 $. 11 and 8
Write the next two terms of $ 80, 40, 20, \dots $
Show the full solution

$ r = \frac{1}{2} $. 10 and 5
Find the 10th term of an arithmetic sequence with $ a_1 = 5 $ and $ d = 3 $.
Show the full solution

$ 5 + (10-1)3 = 5 + 27 = 32 $. Nine gaps, not ten. 32
Find the 6th term of a geometric sequence with $ a_1 = 2 $ and $ r = 3 $.
Show the full solution

$ 2 \cdot 3^5 = 2 \times 243 = 486 $. 486
Write the explicit rule for $ 4, 9, 14, 19, \dots $
Show the full solution

$ d = 5 $, $ a_1 = 4 $: $ a_n = 4 + (n-1)5 = 5n - 1 $. Check $ n=3 $: $ 14 $ ✓. $ a_n = 5n - 1 $
Write the explicit rule for $ 5, 15, 45, 135, \dots $
Show the full solution

$ r = 3 $, $ a_1 = 5 $. $ a_n = 5 \cdot 3^{\,n-1} $
Write a recursive rule for $ 100, 93, 86, \dots $
Show the full solution

$ d = -7 $. The rule needs the starting value too. $ a_1 = 100 $, $ a_n = a_{n-1} - 7 $
Write a recursive rule for $ 6, 18, 54, \dots $
Show the full solution

$ r = 3 $. $ a_1 = 6 $, $ a_n = 3a_{n-1} $
An arithmetic sequence has $ a_1 = 12 $ and $ a_5 = 32 $. Find $ d $.
Show the full solution

Four gaps between the 1st and 5th terms, and the value rose by 20, so $ d = 20 \div 4 = 5 $. $ d = 5 $
An arithmetic sequence has $ a_3 = 11 $ and $ a_8 = 31 $. Find the explicit rule.
Show the full solution

Five gaps, rise of 20, so $ d = 4 $. Back three gaps from $ a_3 $: $ a_1 = 11 - 3(4) = -1 $. Rule: $ a_n = -1 + (n-1)4 = 4n - 5 $. Check $ n=8 $: $ 27 $… recompute: $ 4(8) - 5 = 27 $, but we need 31. The error is the back-step: from $ a_3 $ to $ a_1 $ is two gaps, not three. $ a_1 = 11 - 2(4) = 3 $, giving $ a_n = 4n - 1 $. Check: $ a_3 = 11 $ ✓, $ a_8 = 31 $ ✓. $ a_n = 4n - 1 $
A theater has 18 seats in row 1 and 2 more in each row after. How many in row 15?
Show the full solution

Arithmetic with $ a_1 = 18 $, $ d = 2 $: $ 18 + 14(2) = 46 $. 46 seats
A bacterial culture doubles hourly, starting at 300. How many after 6 hours?
Show the full solution

Geometric with $ r = 2 $. Careful with indexing: 300 is the count at hour 0, so after 6 hours it has doubled six times: $ 300 \times 2^6 = 19200 $. 19 200
A student finds the 12th term of $ 4, 7, 10, \dots $ as $ 4 + 12(3) = 40 $. Find the error.
Show the full solution

They used 12 gaps instead of 11. Twelve terms have eleven gaps between them: $ 4 + 11(3) = 37 $. Listing a few terms confirms the pattern, the $ n $th term is $ 3n + 1 $, so the 12th is 37. 37

Topic 4.2 · Unit 4 · CA CCSS F-LE.1–2, F-IF.7e

Exponential growth and decay

Every exponential model in this course is $ y = a \cdot b^{\,x} $: $ a $ is where you start, $ b $ is what you multiply by each period. Most lost marks come from building $ b $ wrongly out of a percentage.

The method

$ a $ is the initial amount: the value when $ x = 0 $, since $ b^0 = 1 $.
$ b $ is the multiplier per period. Build it from a percentage: growth of $ r\% $ gives $ b = 1 + \frac{r}{100} $; decay of $ r\% $ gives $ b = 1 - \frac{r}{100} $. A 7% rise means $ b = 1.07 $, not 0.07; you keep the original amount and add 7%.
Read $ b $ backwards too. $ b = 1.25 $ is 25% growth; $ b = 0.85 $ is 15% decay, because you keep 85%. $ b > 1 $ grows, $ 0 < b < 1 $ decays.
Watch the period. If $ x $ is in years but the rate is monthly, either convert the rate or redefine $ x $. Mismatching these is a silent error the numbers will not flag.
The graph has a horizontal asymptote at $ y = 0 $; it approaches the axis without touching. Growth rises steeply to the right; decay falls toward the axis.

Where marks are lost: writing $ y = 500(0.03)^x $ for "3% growth". That model multiplies by 0.03 each year, destroying 97% of the value annually. The correct multiplier is 1.03. Sanity-check by computing one period by hand.

Worked examples

Example 1: building the model from a percentage. A $2000 investment grows 6% per year. Write the model and find its value after 10 years.

$ a = 2000 $ and $ b = 1 + 0.06 = 1.06 $, so $ V = 2000(1.06)^t $.

$ V(10) = 2000(1.06)^{10} = 2000 \times 1.7908 \approx \$3581.70 $. Check the direction: the money grew, as 6% growth requires.

Example 2: decay. A car worth $24 000 loses 15% of its value each year. Write the model and find its value after 4 years.

Losing 15% means keeping 85%, so $ b = 0.85 $ and $ V = 24000(0.85)^t $.

$ V(4) = 24000(0.85)^4 = 24000 \times 0.52200625 \approx \$12528 $. Note this is not $ 24000 - 4(15\%) $, each year's loss is taken from a smaller amount, which is exactly what makes it exponential rather than linear.

Example 3: reading a model backwards. A population is modeled by $ P = 4500(0.92)^t $. Describe what is happening.

$ a = 4500 $ is the starting population. $ b = 0.92 $ is less than 1, so this is decay, and $ 1 - 0.92 = 0.08 $ means it falls by 8% per year. Each year 92% of the population remains.

Practice · 15 problems

1–6 build or read models, 7–11 evaluate them, 12–15 apply and diagnose.

Write the multiplier $ b $ for 5% growth.
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$ 1 + 0.05 $. $ b = 1.05 $
Write the multiplier $ b $ for 20% decay.
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Losing 20% means keeping 80%. $ b = 0.80 $
Does $ y = 300(1.12)^x $ grow or decay, and at what rate?
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$ b > 1 $, so growth, at $ 1.12 - 1 = 0.12 $. Growth, 12% per period
Does $ y = 80(0.75)^x $ grow or decay, and at what rate?
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$ b < 1 $, so decay. Keeping 75% means losing 25%. Decay, 25% per period
Write a model for a $500 deposit growing 4% a year.
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$ V = 500(1.04)^t $
Write a model for a 900 mg dose decaying 30% each hour.
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Keeping 70%. $ A = 900(0.70)^h $
For $ y = 200(1.5)^x $, find $ y $ when $ x = 3 $.
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$ 1.5^3 = 3.375 $, so $ 200 \times 3.375 = 675 $. 675
For $ y = 640(0.5)^x $, find $ y $ when $ x = 4 $.
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$ 0.5^4 = 0.0625 $, so $ 640 \times 0.0625 = 40 $. Halving four times: 320, 160, 80, 40 ✓. 40
For $ y = 50(1.2)^x $, find the initial value and the value after 2 periods.
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Initial is 50 at $ x = 0 $. $ 1.2^2 = 1.44 $, so $ 50 \times 1.44 = 72 $. 50 and 72
A town of 8000 grows 3% a year. Find the population after 5 years.
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$ 8000(1.03)^5 = 8000 \times 1.159274 \approx 9274 $. about 9274
A $1200 laptop loses 25% of its value yearly. What is it worth after 3 years?
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$ 1200(0.75)^3 = 1200 \times 0.421875 = 506.25 $. $506.25
A culture triples every hour from 40 cells. Write a model and find the count after 4 hours.
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Tripling means $ b = 3 $: $ N = 40(3)^h $. $ 40 \times 81 = 3240 $. $ N = 40(3)^h $; 3240 cells
An investment doubles every 7 years. Write a model in terms of years $ t $.
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The doubling happens once per 7 years, so the exponent must count seven-year blocks: $ V = a \cdot 2^{\,t/7} $. Writing $ 2^t $ would double every year instead. $ V = a \cdot 2^{\,t/7} $
A student models 3% annual growth on $500 as $ y = 500(0.03)^t $. Find the error.
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Their multiplier destroys 97% of the value each year, after one year it gives $15. Growth of 3% keeps the original and adds 3%, so $ b = 1.03 $: $ y = 500(1.03)^t $, giving $515 after one year. $ y = 500(1.03)^t $
A student says a car losing 15% a year for 4 years has lost 60% of its value. Is that right?
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No; that treats it as linear. Each year's loss comes off a smaller amount: $ 0.85^4 \approx 0.522 $, so about 52% of the value remains and roughly 48% has been lost, not 60%. About 48% lost

Topic 4.3 · Unit 4 · CA CCSS F-LE.1, F-LE.3

Comparing linear with exponential

The test is one question: does the quantity change by a constant amount or a constant factor? Adding the same number each step is linear; multiplying by the same number is exponential. Everything else in this topic follows from that distinction.

The method

From a table: check the $ x $-values are equally spaced, then subtract consecutive $ y $-values. Constant differences → linear. If not, divide them. Constant ratios → exponential.
From words: "per", "each", "every" followed by a fixed amount signals linear. A percentage, or "doubles", "halves", "triples", signals exponential.
From a graph: a line is straight; an exponential curves, steepening for growth or flattening toward an asymptote for decay.
Growth rates: an exponential with $ b > 1 $ will eventually exceed any linear function, however small its rate and however steep the line. The crossing may be far to the right, but it always happens.
Watch the starting point. A linear model often leads early on and gets overtaken later, which is exactly what comparison questions test.

Where marks are lost: deciding from the first two rows alone. $ 2, 4, \dots $ could be adding 2 or multiplying by 2; you need a third term to tell $ 2, 4, 6 $ (linear) from $ 2, 4, 8 $ (exponential).

Worked examples

Example 1: from a table. Classify $ (0, 3), (1, 6), (2, 12), (3, 24) $.

Differences: 3, 6, 12, not constant, so not linear. Ratios: $ 6 \div 3 = 2 $, $ 12 \div 6 = 2 $, $ 24 \div 12 = 2 $, constant. Exponential, with $ a = 3 $ and $ b = 2 $, so $ y = 3(2)^x $.

Example 2: from a description. Which is which? (i) A tree 4 m tall grows 0.5 m each year. (ii) A tree 4 m tall grows 5% each year.

(i) adds a fixed amount, so linear: $ h = 4 + 0.5t $. (ii) multiplies by a fixed factor, so exponential: $ h = 4(1.05)^t $. The wording "each year" appears in both, what distinguishes them is whether what follows is an amount or a percentage.

Example 3: the overtake. Job A pays $40 000 with a $2000 rise each year. Job B pays $35 000 with a 5% rise each year. Which pays more in year 10, and in year 20?

A: $ 40000 + 2000t $. B: $ 35000(1.05)^t $.

At $ t = 10 $: A gives $60 000; B gives $ 35000 \times 1.6289 \approx \$57{,}011 $. A leads. At $ t = 20 $: A gives $80 000; B gives $ 35000 \times 2.6533 \approx \$92{,}866 $. B has overtaken. The exponential started lower and grew more slowly in absolute terms at first, but the constant factor always wins eventually.

Practice · 15 problems

1–6 classify, 7–11 build the matching model, 12–15 compare and diagnose.

Classify: $ (0, 5), (1, 8), (2, 11), (3, 14) $.
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Differences all 3. Linear
Classify: $ (0, 2), (1, 6), (2, 18), (3, 54) $.
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Ratios all 3. Exponential
Classify: $ (0, 100), (1, 90), (2, 81), (3, 72.9) $.
Show the full solution

Differences are $ -10, -9, -8.1 $, not constant. Ratios are all 0.9. Exponential decay
Classify: a phone plan charging $20 plus $0.05 per text.
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A fixed amount per text is added. Linear
Classify: a savings account paying 2% interest a year.
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A percentage means multiplying by a fixed factor. Exponential
Classify: a pool losing 40 liters an hour.
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A fixed amount subtracted each hour. Linear (decreasing)
Write the model for $ (0, 7), (1, 10), (2, 13) $.
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Linear, slope 3, intercept 7. $ y = 3x + 7 $
Write the model for $ (0, 4), (1, 12), (2, 36) $.
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Exponential, $ a = 4 $, $ r = 3 $. $ y = 4(3)^x $
Write the model for $ (0, 64), (1, 32), (2, 16) $.
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Exponential with $ r = 0.5 $. $ y = 64(0.5)^x $
A membership costs $50 to join plus $15 a month. Write the model and classify it.
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$ C = 50 + 15m $, linear
A $6000 car loses 12% a year. Write the model and classify it.
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$ V = 6000(0.88)^t $, exponential decay
Company A earns $500 a week. Company B earns $100 in week 1 and doubles weekly. Who has earned more in week 4, and in week 6?
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A: 500 each week. B: 100, 200, 400, 800 in weeks 1–4, so B earns 800 in week 4, already ahead. By week 6 B earns $ 100 \times 2^5 = 3200 $. B in both, from week 4 onward
A linear model gives $ y = 1000 + 100x $ and an exponential gives $ y = 100(1.5)^x $. Which is larger at $ x = 5 $, and what does that suggest?
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Linear: $ 1000 + 500 = 1500 $. Exponential: $ 100 \times 7.59 \approx 759 $. The linear leads at $ x = 5 $, but since the exponential has $ b > 1 $ it must overtake eventually, starting lower only delays the crossing. Linear now; exponential later
A student sees $ 3, 9, \dots $ and concludes the sequence is exponential. Why is that premature?
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Two terms cannot distinguish adding 6 from multiplying by 3. $ 3, 9, 15 $ is linear; $ 3, 9, 27 $ is exponential. A third term is required. Need a third term
A student says a linear function with a large slope will always beat an exponential with a small growth rate. Is that right?
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No. Any exponential with $ b > 1 $ eventually exceeds any linear function, however steep. A large slope and a small rate only push the crossing point further right; they do not prevent it. False

Topic 4.4 · Unit 4 · CA CCSS F-BF.3

Transformations of functions

One set of rules governs every parent function you meet this year. Changes outside the function move it vertically and behave as you expect; changes inside move it horizontally and behave backwards.

The method

Starting from $ f(x) $:

$ f(x) + k $: shifts up $ k $. Negative $ k $ shifts down. Outside, so intuitive.
$ f(x - h) $: shifts right $ h $. Inside, so backwards: $ f(x - 3) $ moves right 3, and $ f(x + 3) $ moves left 3.
$ a \cdot f(x) $: stretches vertically by $ a $. If $ |a| > 1 $ it is steeper; if $ 0 < |a| < 1 $ it is flatter.
$ -f(x) $: reflects across the $ x $-axis (flips upside down). $ f(-x) $: reflects across the $ y $-axis.
Order matters when combining: apply horizontal shifts and stretches first, then reflections, then vertical shifts last.

Why the horizontal rule is backwards: in $ f(x - 3) $ the function needs its input to equal the old value, so $ x $ must be 3 larger to produce the same output, every point moves right. Checking one point beats memorizing the rule.

Worked examples

Example 1: describing a combination. Describe $ g(x) = 2|x + 3| - 5 $ as transformations of $ f(x) = |x| $.

Inside: $ x + 3 = x - (-3) $, so shift left 3. Multiplier 2 outside: vertical stretch by 2. Then $ -5 $: shift down 5. The vertex moves from $ (0,0) $ to $ (-3, -5) $, and the arms are twice as steep.

Example 2: a reflection. Describe $ g(x) = -(x - 2)^2 + 4 $ from $ f(x) = x^2 $.

$ (x - 2) $: right 2. The leading minus: reflect across the $ x $-axis, so the parabola opens downward. Then $ +4 $: up 4. The vertex sits at $ (2, 4) $ and is a maximum rather than a minimum because of the reflection.

Example 3: writing the rule from a description. Take $ f(x) = 2^x $, shift it right 1 and down 3. Write $ g(x) $.

Right 1 goes inside as $ x - 1 $; down 3 goes outside as $ -3 $: $ g(x) = 2^{\,x-1} - 3 $.

Check a point: $ f(0) = 1 $, so after moving right 1 and down 3 that point should land at $ (1, -2) $. Test: $ g(1) = 2^0 - 3 = -2 $ ✓.

Practice · 15 problems

1–6 single transformations, 7–11 combinations, 12–15 write rules and diagnose.

Describe $ f(x) + 7 $.
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Outside the function. Shift up 7
Describe $ f(x) - 2 $.
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Shift down 2
Describe $ f(x - 5) $.
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Inside, so backwards from the sign. Shift right 5
Describe $ f(x + 4) $.
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Shift left 4
Describe $ -f(x) $.
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The output is negated. Reflection across the $ x $-axis
Describe $ 3f(x) $.
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Vertical stretch by 3
Describe $ f(x - 2) + 6 $.
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Right 2 and up 6
Describe $ -2f(x) + 1 $.
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Stretch by 2, reflect across the $ x $-axis, then shift up 1. Stretch 2, reflect, up 1
Give the vertex of $ y = |x - 4| + 2 $.
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Right 4 and up 2 from the origin. $ (4, 2) $
Give the vertex of $ y = -(x + 1)^2 - 3 $.
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$ x + 1 $ means left 1; $ -3 $ means down 3; the minus makes it open downward. $ (-1, -3) $, a maximum
Describe $ y = 2^{\,x} + 5 $ from $ y = 2^x $, and state its asymptote.
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Shift up 5. The asymptote moves up with the graph, from $ y = 0 $ to $ y = 5 $. Up 5; asymptote $ y = 5 $
Write the rule for $ f(x) = x^2 $ shifted left 3 and up 1.
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Left 3 is $ x + 3 $ inside; up 1 is $ +1 $ outside. $ y = (x + 3)^2 + 1 $
Write the rule for $ f(x) = |x| $ reflected across the $ x $-axis and shifted down 4.
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$ y = -|x| - 4 $
The point $ (2, 5) $ is on $ y = f(x) $. Where does it go on $ y = f(x - 3) + 2 $?
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Right 3 and up 2. $ (5, 7) $
A student says $ y = (x + 6)^2 $ shifts the parabola right 6. Find the error.
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Horizontal shifts work backwards from the sign. Matching the form $ (x - h)^2 $ gives $ h = -6 $, a shift left 6. Check: the vertex is where the bracket is zero, at $ x = -6 $. Left 6

Unit 4 mixed review · 10 problems · all topics

Unit 4 mixed review: Sequences and Exponentials

Sequences, exponential models, linear-versus-exponential decisions, and transformations.

Find the 12th term of an arithmetic sequence with $ a_1 = 6 $ and $ d = 4 $.
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$ 6 + 11(4) = 50 $. Eleven gaps, not twelve. 50
Write the explicit rule for $ 2, 10, 50, 250, \dots $
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Ratios all 5. $ a_n = 2 \cdot 5^{\,n-1} $
Write a model for $800 growing 7% a year.
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$ V = 800(1.07)^t $
Does $ y = 120(0.85)^x $ grow or decay, and at what rate?
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$ b < 1 $, so decay; keeping 85% means losing 15%. Decay, 15%
Classify $ (0, 6), (1, 10), (2, 14), (3, 18) $.
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Differences all 4. Linear
Classify $ (0, 5), (1, 10), (2, 20), (3, 40) $ and write its model.
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Ratios all 2. Exponential, $ y = 5(2)^x $
Describe $ y = -|x + 2| + 3 $ as transformations of $ y = |x| $.
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Left 2, reflect across the $ x $-axis, up 3. The vertex is $ (-2, 3) $ and it is a maximum. Left 2, reflect, up 3
A $900 phone loses 20% of its value yearly. What is it worth after 3 years?
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$ 900(0.8)^3 = 900 \times 0.512 = 460.80 $. $460.80
An arithmetic sequence has $ a_2 = 9 $ and $ a_6 = 29 $. Find $ d $ and $ a_1 $.
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Four gaps, rise of 20, so $ d = 5 $. One gap back from $ a_2 $: $ a_1 = 9 - 5 = 4 $. $ d = 5 $, $ a_1 = 4 $
The point $ (3, 4) $ lies on $ y = f(x) $. Where is it on $ y = f(x + 1) - 6 $?
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$ x + 1 $ shifts left 1; $ -6 $ shifts down 6. $ (2, -2) $

Topic 5.1 · Unit 5 · CA CCSS S-ID.1–3

One-variable data: center, spread and outliers

Two numbers describe a data set: one for where it sits and one for how spread out it is. The skill being tested is choosing the right pair, because an outlier wrecks the mean and leaves the median untouched.

The method

Order the data first. Almost every error in this topic traces back to finding a median without sorting.
Center. Mean is the total divided by the count. Median is the middle value, or the mean of the two middle values when the count is even.
Spread. Range is max minus min. The interquartile range is $ Q_3 - Q_1 $, where $ Q_1 $ is the median of the lower half and $ Q_3 $ the median of the upper half. Standard deviation measures typical distance from the mean.
Choose the pair to match the shape. Roughly symmetric with no outliers → mean and standard deviation. Skewed, or containing an outlier → median and IQR, because both ignore extreme values.
The outlier test: anything below $ Q_1 - 1.5 \times \text{IQR} $ or above $ Q_3 + 1.5 \times \text{IQR} $. Quote the test, not a hunch, when a question asks you to justify.

Where marks are lost: including the median itself when splitting the data into halves for the quartiles. With an odd count, leave the median out of both halves. Also, "the mean went up so the data improved" is not a safe reading, one large value can lift a mean while most of the data falls.

Worked examples

Example 1: the effect of an outlier. Find the mean and median of $ 4, 5, 6, 7, 8 $, then of $ 4, 5, 6, 7, 68 $.

First set: mean $ = 30 \div 5 = 6 $; median $ = 6 $. They agree, which suggests symmetry.

Second set: mean $ = 90 \div 5 = 18 $; median $ = 6 $. One value dragged the mean to 18, a figure larger than four of the five data points, so the mean now describes nothing in the set. The median did not move at all. With an outlier present, report the median.

Example 2: quartiles and IQR. Find $ Q_1 $, $ Q_3 $ and the IQR of $ 3, 7, 8, 12, 14, 18, 21 $.

Seven values, so the median is the 4th: 12. Now exclude it. Lower half: $ 3, 7, 8 $ → $ Q_1 = 7 $. Upper half: $ 14, 18, 21 $ → $ Q_3 = 18 $. IQR $ = 18 - 7 = 11 $.

Example 3: testing for an outlier. Using those quartiles, is 45 an outlier if added to the set?

$ 1.5 \times \text{IQR} = 16.5 $. Upper fence: $ Q_3 + 16.5 = 18 + 16.5 = 34.5 $. Since $ 45 > 34.5 $, yes, 45 is an outlier by the standard test. Stating the fence is what earns the mark; saying "45 looks too big" does not.

Practice · 15 problems

1–6 compute, 7–11 involve quartiles and outliers, 12–15 interpret and diagnose.

Find the mean of $ 6, 9, 12, 13 $.
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$ 40 \div 4 = 10 $. 10
Find the median of $ 11, 4, 9, 2, 7 $.
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Sort first: $ 2, 4, 7, 9, 11 $. The middle of five is the 3rd. 7
Find the median of $ 3, 8, 10, 15 $.
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Even count, so average the two middle values: $ (8 + 10) \div 2 = 9 $. 9
Find the range of $ 14, 6, 22, 9 $.
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$ 22 - 6 = 16 $. 16
Find the mean of $ 5, 5, 5, 25 $ and say whether it represents the data well.
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$ 40 \div 4 = 10 $. But three of the four values are 5, so a "typical" value of 10 misleads, 25 pulled it up. The median, 5, describes the set better. Mean 10; median is more representative
Find $ Q_1 $, $ Q_3 $ and the IQR of $ 2, 4, 6, 8, 10, 12 $.
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Even count, so the halves are $ 2,4,6 $ and $ 8,10,12 $. $ Q_1 = 4 $, $ Q_3 = 10 $, IQR $ = 6 $. 4, 10, 6
Find $ Q_1 $, $ Q_3 $ and the IQR of $ 5, 7, 9, 11, 13, 15, 17 $.
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Median is 11; exclude it. Lower: $ 5,7,9 $ → $ Q_1 = 7 $. Upper: $ 13,15,17 $ → $ Q_3 = 15 $. IQR $ = 8 $. 7, 15, 8
A data set has $ Q_1 = 20 $ and $ Q_3 = 32 $. What are the outlier fences?
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IQR $ = 12 $, so $ 1.5 \times 12 = 18 $. Lower fence $ 20 - 18 = 2 $; upper fence $ 32 + 18 = 50 $. Below 2 or above 50
Using those fences, is 51 an outlier? Is 45?
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$ 51 > 50 $, so yes. $ 45 < 50 $, so no, even though it is well above $ Q_3 $, the test is the fence, not the impression. 51 yes; 45 no
Which measure of center should you report for house prices in a city, and why?
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The median. House price data is strongly right-skewed, a few very expensive homes pull the mean far above what a typical buyer pays. Median, because of right skew
A set has mean 50 and median 50. What does that suggest about its shape?
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They agree, which is consistent with a roughly symmetric distribution and no strong outliers. It is evidence rather than proof, unusual sets can match means and medians without being symmetric. Likely roughly symmetric
A set has mean 62 and median 45. What does that suggest?
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The mean sits well above the median, so high values are pulling it up. Right-skewed, probably with high outliers
Test scores are $ 70, 72, 74, 75, 78 $. Every student then receives 5 bonus marks. What happens to the mean and to the range?
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Adding a constant shifts every value, so the mean rises by 5. The gaps between values are unchanged, so the range stays the same. Mean +5; range unchanged
A student finds the median of $ 9, 3, 7, 1, 5 $ by taking the middle of the list as written and answers 7. Find the error.
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They did not sort. In order the set is $ 1, 3, 5, 7, 9 $, so the median is 5. The middle position only gives the median once the data is ordered. 5
A student computes the IQR of $ 4, 6, 9, 11, 14 $ by splitting into $ 4,6,9 $ and $ 9,11,14 $, getting $ Q_1 = 6 $, $ Q_3 = 11 $. What did they do wrong?
Show the full solution

They included the median, 9, in both halves. With an odd count it belongs to neither: the halves are $ 4,6 $ and $ 11,14 $, giving $ Q_1 = 5 $, $ Q_3 = 12.5 $ and IQR $ = 7.5 $. IQR $ = 7.5 $

Topic 5.2 · Unit 5 · CA CCSS S-ID.5

Two-way frequency tables

A two-way table counts people by two categories at once. Every question reduces to one decision: what is the denominator? Get that right and the arithmetic is trivial; get it wrong and a correct calculation still earns nothing.

The method

Take a table of students by year group and whether they play a sport, with a grand total of 200.

Joint relative frequency: a cell divided by the grand total. "What fraction of all students are juniors who play a sport?"
Marginal relative frequency: a row or column total divided by the grand total. "What fraction of all students are juniors?"
Conditional relative frequency: a cell divided by its row or column total. "Of the juniors, what fraction play a sport?" The words "of the", "given that" and "among" all signal a conditional.
Read the question for the denominator. "What percentage of athletes are juniors" and "what percentage of juniors are athletes" are different questions with different denominators and usually different answers.
To judge association: compare conditional frequencies across groups. If 60% of juniors play a sport but only 35% of seniors do, year group and participation appear associated.

Where marks are lost: dividing by the grand total when the question said "of the juniors". If the question restricts attention to a group, that group's total is the denominator, the rest of the table is irrelevant.

Worked examples

The table for all three examples. Of 200 students: 80 juniors, of whom 48 play a sport; 120 seniors, of whom 42 play a sport. So 90 students play a sport and 110 do not.

Example 1: joint. What fraction of all students are juniors who play a sport?

The cell is 48 and the denominator is the grand total: $ 48 \div 200 = 0.24 $, or 24%.

Example 2: conditional, one way. Of the juniors, what percentage play a sport?

"Of the juniors" restricts to that row, so the denominator is 80: $ 48 \div 80 = 0.60 $, or 60%.

Example 3: conditional, the other way. Of the students who play a sport, what percentage are juniors?

Now the restriction is to athletes, so the denominator is 90: $ 48 \div 90 \approx 0.533 $, or about 53.3%. Same cell, different denominator, and a genuinely different answer; this reversal is the single most tested idea in the topic.

Practice · 15 problems

Problems 1–10 use this table: 150 people surveyed, 90 own a pet, 60 do not. Of the pet owners, 54 live in a house and 36 in a flat. Of the non-owners, 24 live in a house and 36 in a flat.

How many people live in a house?
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$ 54 + 24 = 78 $. 78
How many people live in a flat?
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$ 36 + 36 = 72 $. Check: $ 78 + 72 = 150 $ ✓. 72
What fraction of all people own a pet and live in a house?
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Joint, so divide by the grand total: $ 54 \div 150 = 0.36 $. 36%
What fraction of all people own a pet?
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Marginal: $ 90 \div 150 = 0.60 $. 60%
Of the pet owners, what percentage live in a house?
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"Of the pet owners" makes 90 the denominator: $ 54 \div 90 = 0.60 $. 60%
Of the people living in a house, what percentage own a pet?
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Denominator is now 78: $ 54 \div 78 \approx 0.692 $. about 69.2%
Of the non-owners, what percentage live in a flat?
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$ 36 \div 60 = 0.60 $. 60%
Of the flat-dwellers, what percentage own a pet?
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$ 36 \div 72 = 0.50 $. 50%
Compare the percentage of house-dwellers owning a pet with the percentage of flat-dwellers owning a pet. Does housing appear associated with pet ownership?
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House: $ 54 \div 78 \approx 69.2\% $. Flat: $ 36 \div 72 = 50\% $. The conditional frequencies differ by about 19 points, which suggests an association. Yes, apparently associated
What fraction of all people neither own a pet nor live in a house?
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That is the non-owner flat cell, 36, over the grand total: $ 36 \div 150 = 0.24 $. 24%
In a different survey, 40% of all respondents are students and 25% of all respondents are students who commute. Of the students, what percentage commute?
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Restricting to students makes 40% the denominator: $ 25 \div 40 = 0.625 $. 62.5%
Explain in one sentence why "the percentage of athletes who are juniors" and "the percentage of juniors who are athletes" usually differ.
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They share a numerator but have different denominators, the first divides by the number of athletes, the second by the number of juniors, so they only coincide when those two totals happen to be equal. Different denominators
A table shows 30 of 50 men and 45 of 75 women prefer option A. Is preference associated with gender?
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Men: $ 30 \div 50 = 60\% $. Women: $ 45 \div 75 = 60\% $. The conditional frequencies are identical, so there is no evidence of association here, despite the raw counts differing. No association
A student is asked "of the 90 pet owners, what percentage live in a flat?" and computes $ 36 \div 150 $. Find the error.
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They used the grand total for a conditional question. "Of the 90 pet owners" names the denominator explicitly: $ 36 \div 90 = 40\% $, not 24%. 40%
A student concludes from the table that owning a pet causes people to live in houses. Why is that unjustified?
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The table shows an association, not a cause. A third factor (income, family size, having a garden) could drive both, and the causation could equally run the other way, with house-dwellers being freer to get a pet. Association is not causation

Topic 5.3 · Unit 5 · CA CCSS S-ID.6–9

Scatter plots, lines of best fit and correlation

This topic connects Unit 2 to data: fit a line to a cloud of points, then use its slope and intercept to say something about the real situation. The two places marks are lost are interpreting the numbers without units, and confusing correlation with causation.

The method

Describe the pattern in three words: form (linear or not), direction (positive or negative), strength (strong, moderate, weak). A question asking you to "describe the relationship" wants all three.
Fit a line that follows the trend with roughly as many points above as below. The line of best fit always passes through the point $ (\bar{x}, \bar{y}) $.
Interpret the slope in context as "output units per input unit", the predicted change in $ y $ for a one-unit rise in $ x $.
Interpret the intercept carefully. It is the predicted $ y $ when $ x = 0 $, which is often meaningless, a newborn's predicted height, a car with zero engine size. Say so when it is.
Correlation coefficient $ r $ runs from $ -1 $ to 1. The sign gives direction, the size gives strength, and $ r $ near 0 means no linear relationship; there could still be a strong curved one.
Residual $ = $ actual $ - $ predicted. Positive means the point sits above the line. A residual plot with no pattern supports a linear model; a curved residual plot says a line was the wrong choice.

Where marks are lost: concluding causation. Ice cream sales and drowning deaths correlate strongly because both rise with temperature, a lurking variable. Only a controlled experiment supports a causal claim.

Worked examples

Example 1: interpreting slope and intercept. For house size $ x $ in hundreds of square feet and price $ y $ in thousands of dollars, the line of best fit is $ \hat{y} = 32x + 85 $. Interpret both numbers.

Slope: each additional hundred square feet is associated with a predicted price rise of $32 000. Intercept: a house of zero square feet is predicted to cost $85 000, which is meaningless, since no such house exists. The intercept here is a mathematical artefact of the fit, not a real prediction, and saying so earns the mark.

Example 2: prediction and residual. Using that line, predict the price of a 1800 square foot house. If it actually sold for $690 000, find the residual.

$ x = 18 $ (hundreds), so $ \hat{y} = 32(18) + 85 = 661 $, predicting $661 000.

Residual $ = 690 - 661 = 29 $, so the house sold for $29 000 more than predicted, and the point sits above the line.

Example 3: correlation without causation. A study finds $ r = 0.87 $ between the number of firefighters at a blaze and the damage caused. Does sending more firefighters cause more damage?

No. The lurking variable is the size of the fire: large fires attract more firefighters and also cause more damage. The correlation is real and strong, and the causal claim does not follow from it. Recommending fewer firefighters would be the error the statistic invites.

Practice · 15 problems

1–5 describe and read, 6–11 predict and compute residuals, 12–15 interpret and diagnose.

A scatter plot shows points rising steadily from lower left to upper right, close to a straight line. Describe the relationship.
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All three words are needed. Strong positive linear
What does $ r = -0.91 $ tell you?
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Negative direction and close to $ -1 $, so a strong negative linear relationship. Strong negative linear
What does $ r = 0.06 $ tell you?
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Almost no linear relationship, though a strong curved one could still exist, which $ r $ would not detect. No linear relationship
For $ \hat{y} = 4x + 20 $, predict $ y $ when $ x = 7 $.
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$ 4(7) + 20 = 48 $. 48
For $ \hat{y} = -3x + 50 $, predict $ y $ when $ x = 12 $.
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$ -36 + 50 = 14 $. 14
A point has actual value 30 and predicted value 26. Find the residual and say where the point sits.
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$ 30 - 26 = 4 $, positive, so the point is above the line. $ +4 $, above
A point has actual value 18 and predicted value 25. Find the residual.
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$ 18 - 25 = -7 $, so the point lies below the line. $ -7 $, below
For study hours $ x $ and test score $ y $, $ \hat{y} = 6x + 52 $. Interpret the slope.
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Each additional hour of study is associated with a predicted 6-point rise in score. Units matter. 6 points per hour
Using that same line, interpret the intercept and say whether it is meaningful.
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A student studying zero hours is predicted to score 52. This one is meaningful, because studying zero hours is a real possibility, unlike a house of zero square feet. 52 points; meaningful here
Using $ \hat{y} = 6x + 52 $, a student studies 5 hours and scores 90. Find the residual.
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Predicted $ 6(5) + 52 = 82 $. Residual $ = 90 - 82 = 8 $. $ +8 $
A residual plot shows a clear U-shape. What does that indicate?
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A pattern in the residuals means the line has missed structure in the data, the relationship is curved, so a linear model is the wrong choice even if $ r $ looks respectable. Linear model is inappropriate
A study finds a strong positive correlation between shoe size and reading ability in children. Explain.
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Age is the lurking variable: older children have bigger feet and read better. Shoe size does not affect reading. Lurking variable: age
Using $ \hat{y} = 6x + 52 $, a student predicts the score for 30 hours of study as 232. What is wrong with that?
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Two problems. It extrapolates far outside the data the line was fitted to, where the linear pattern may not hold; and a test score above 100 is impossible, which exposes the model's limits. Invalid extrapolation
A student reports $ r = 1.4 $ for a data set. What does that tell you?
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$ r $ is always between $ -1 $ and 1, so a value of 1.4 is impossible and indicates a calculation error. Impossible value
A newspaper reports that students who eat breakfast score higher, and concludes that eating breakfast raises grades. Why is that not supported?
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This is observational data showing an association. Households where breakfast is routine may differ in income, sleep, or supervision, any of which could drive both. Only a controlled experiment assigning breakfast at random would support the causal claim. Observational; causation not supported

Unit 5 mixed review · 10 problems · all topics

Unit 5 mixed review: Descriptive Statistics

One-variable summaries, two-way tables, and scatter plots.

Find the median of $ 12, 5, 19, 8, 14 $.
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Sort: $ 5, 8, 12, 14, 19 $. 12
Find the mean of $ 4, 8, 9, 15 $.
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$ 36 \div 4 = 9 $. 9
A set has $ Q_1 = 14 $ and $ Q_3 = 26 $. Find the upper outlier fence.
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IQR $ = 12 $, so $ 1.5 \times 12 = 18 $ and the fence is $ 26 + 18 = 44 $. 44
A set has mean 70 and median 55. What does that suggest?
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The mean is well above the median, so high values are pulling it up. Right-skewed
Of 200 people, 120 are students and 72 of those students own a bike. Of the students, what percentage own a bike?
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"Of the students" makes 120 the denominator: $ 72 \div 120 = 0.60 $. 60%
Using that data, what fraction of all 200 people are students who own a bike?
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Joint, so divide by the grand total: $ 72 \div 200 = 0.36 $. 36%
What does $ r = -0.88 $ tell you?
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Strong negative linear relationship
For $ \hat{y} = 7x + 30 $, predict $ y $ at $ x = 6 $ and find the residual if the actual value is 65.
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Predicted $ 42 + 30 = 72 $. Residual $ = 65 - 72 = -7 $, so the point sits below the line. Predicted 72; residual $ -7 $
A study finds a strong correlation between the number of umbrellas sold and traffic accidents. Does one cause the other?
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No, rainfall is the lurking variable, raising both. No; lurking variable
A student computes the IQR of $ 3, 5, 8, 11, 15 $ by including the median in both halves. What is the correct IQR?
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With an odd count, exclude the median 8. Halves are $ 3,5 $ and $ 11,15 $, so $ Q_1 = 4 $ and $ Q_3 = 13 $, giving IQR $ = 9 $. 9

Topic 6.1 · Unit 6 · CA CCSS G-CO.2–7

Rigid transformations and congruence

California's standards define congruence through motion: two figures are congruent when one can be moved onto the other by translations, reflections and rotations. That definition is what most questions in this unit are really testing, so the coordinate rules are worth knowing cold.

The method

The three rigid motions: rigid because each preserves distance and angle, so the image is always congruent to the original.

Translation $ (x, y) \to (x + a, y + b) $. Slides without turning or flipping.
Reflection. Across the $ x $-axis: $ (x, y) \to (x, -y) $. Across the $ y $-axis: $ (x, y) \to (-x, y) $. Across $ y = x $: $ (x, y) \to (y, x) $. Remember which coordinate changes by asking which axis you are jumping over, reflecting over the $ x $-axis changes height, so $ y $ flips.
Rotation about the origin. 90° counter-clockwise: $ (x, y) \to (-y, x) $. 180°: $ (x, y) \to (-x, -y) $. 270° counter-clockwise: $ (x, y) \to (y, -x) $.
To show two figures are congruent: describe a specific sequence of rigid motions carrying one exactly onto the other. "They look the same" earns nothing; "reflect across the $ y $-axis, then translate 3 down" earns the mark.
Dilation is not rigid. It changes size, so it produces a similar figure, not a congruent one.

Where marks are lost: rotation direction. Unless a question says clockwise, assume counter-clockwise. A 90° clockwise rotation is the same as 270° counter-clockwise, giving $ (x, y) \to (y, -x) $, check by rotating a single point you can picture, such as $ (1, 0) $.

Worked examples

Example 1: applying a rule. Translate $ A(2, -3) $ by $ (x, y) \to (x - 4, y + 5) $.

$ x: 2 - 4 = -2 $; $ y: -3 + 5 = 2 $. So $ A' = (-2, 2) $. The figure moves 4 left and 5 up.

Example 2: describing an unknown transformation. A triangle has vertices $ (1, 2), (4, 2), (1, 6) $. Its image is $ (-1, 2), (-4, 2), (-1, 6) $. Describe the transformation.

Each $ x $-coordinate has changed sign while every $ y $ stayed put. That is exactly $ (x, y) \to (-x, y) $. A reflection across the $ y $-axis. Checking every vertex rather than one is what makes this safe, a single point could match several different transformations.

Example 3: a sequence proving congruence. Show that triangle $ (0,0), (3,0), (0,4) $ is congruent to $ (5,1), (5,4), (1,1) $.

The first has legs 3 and 4 along the axes. The second has a vertical leg from $ (5,1) $ to $ (5,4) $, length 3, and a horizontal leg from $ (5,1) $ to $ (1,1) $, length 4, the legs have swapped orientation, which suggests a rotation.

Rotate the first 90° counter-clockwise about the origin: $ (x,y) \to (-y,x) $ sends $ (0,0) \to (0,0) $, $ (3,0) \to (0,3) $, $ (0,4) \to (-4,0) $. Now translate by $ (+5, +1) $: $ (5,1) $, $ (5,4) $, $ (1,1) $, exactly the target. Rotate 90° counter-clockwise about the origin, then translate 5 right and 1 up.

Practice · 15 problems

1–6 apply rules, 7–11 identify transformations, 12–15 build sequences and diagnose.

Translate $ (3, 5) $ by $ (x, y) \to (x + 2, y - 4) $.
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$ (3+2, 5-4) $. $ (5, 1) $
Reflect $ (6, -2) $ across the $ x $-axis.
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$ (x, -y) $, so the $ y $ sign flips. $ (6, 2) $
Reflect $ (6, -2) $ across the $ y $-axis.
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$ (-x, y) $. $ (-6, -2) $
Rotate $ (4, 1) $ by 180° about the origin.
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$ (-x, -y) $. $ (-4, -1) $
Rotate $ (2, 7) $ by 90° counter-clockwise about the origin.
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$ (-y, x) $. $ (-7, 2) $
Reflect $ (3, 8) $ across the line $ y = x $.
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Swap the coordinates. $ (8, 3) $
$ (5, 2) $ maps to $ (5, -2) $. Name the transformation.
Show the full solution

$ x $ unchanged, $ y $ negated. Reflection across the $ x $-axis
$ (1, 4) $ maps to $ (-4, 1) $. Name the transformation.
Show the full solution

This matches $ (x,y) \to (-y,x) $. 90° counter-clockwise rotation about the origin
$ (2, 3) $ maps to $ (7, 0) $. Name the transformation.
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$ x $ rose by 5 and $ y $ fell by 3, with no sign changes. Translation $ (x+5, y-3) $
A triangle is reflected across the $ y $-axis. Is the image congruent to the original? Why?
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Yes. Reflection is a rigid motion, so distances and angles are preserved, which is precisely the definition of congruent in this course. Yes, reflection is rigid
A triangle is dilated by a factor of 2. Is the image congruent to the original?
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No. Dilation changes the side lengths, so it is not rigid. The image is similar, same shape, different size. No, similar, not congruent
Apply a reflection across the $ x $-axis followed by a translation $ (x+3, y+1) $ to the point $ (2, 5) $.
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Reflect: $ (2, -5) $. Then translate: $ (5, -4) $. Order matters, doing the translation first would give a different result. $ (5, -4) $
Describe a sequence carrying $ (0,0), (2,0), (0,3) $ onto $ (0,0), (-2,0), (0,-3) $.
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Both non-origin vertices have had both coordinates negated, which is $ (x,y) \to (-x,-y) $. 180° rotation about the origin
A student rotates $ (3, 0) $ by 90° counter-clockwise and answers $ (0, -3) $. Find the error.
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They applied the clockwise rule. Counter-clockwise is $ (x,y) \to (-y,x) $, giving $ (0, 3) $. Picture it: a point on the positive $ x $-axis turning counter-clockwise swings up into the positive $ y $-axis. $ (0, 3) $
A student says two triangles are congruent "because they look the same size". Why does that not earn the mark?
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Congruence must be demonstrated, either by a specific sequence of rigid motions carrying one exactly onto the other, or by a congruence criterion such as SSS or SAS. Appearance is not evidence. Needs a sequence or criterion

Topic 6.2 · Unit 6 · CA CCSS G-CO.7–8

Triangle congruence criteria and proof

You do not need all six pairs of corresponding parts to prove two triangles congruent, three of the right kind suffice. Knowing which three, and writing the reason for each line, is what separates a proof that scores from one that does not.

The method

The four criteria that work:

SSS: all three sides.
SAS: two sides and the angle between them.
ASA: two angles and the side between them.
AAS: two angles and a side not between them.

The two that do not: SSA fails, because two different triangles can share the same two sides and non-included angle. AAA fails too; it gives similar triangles of any size, not congruent ones.

Mark the diagram with everything you are given, then add everything free: shared sides, vertical angles, parallel-line angle pairs.
Count what you have and in what order around the triangle. SAS needs the angle between the two sides; if it is not between them you have SSA and no conclusion.
Write two columns: statements on the left, reasons on the right. Every line needs a reason, given, a definition, a theorem, or the reflexive property.
Name the triangles in corresponding order. $ \triangle ABC \cong \triangle DEF $ claims $ A $ matches $ D $, $ B $ matches $ E $, $ C $ matches $ F $. Scrambling the order makes the statement wrong even when the triangles are congruent.
CPCTC (corresponding parts of congruent triangles are congruent) is the reason you use after proving congruence, to conclude something about a remaining side or angle. Never before.

Free facts students forget to use: a shared side is congruent to itself (reflexive property); vertical angles are congruent; and where lines are parallel, alternate interior angles are congruent. Most two-column proofs in this course need exactly one of these to complete the third pair.

Worked examples

Example 1: choosing the criterion. Two triangles share $ AB = DE = 7 $, $ BC = EF = 5 $, and $ \angle B = \angle E = 40° $. Which criterion applies?

$ \angle B $ sits between sides $ AB $ and $ BC $, and $ \angle E $ between $ DE $ and $ EF $. The angle is included, so SAS. Had the given angle been $ \angle A $, it would not lie between the two sides, leaving SSA and no valid conclusion.

Example 2: a proof using the reflexive property. Given $ AB \cong CB $ and $ AD \cong CD $, prove $ \triangle ABD \cong \triangle CBD $.

Statement	Reason
$ AB \cong CB $	Given
$ AD \cong CD $	Given
$ BD \cong BD $	Reflexive property
$ \triangle ABD \cong \triangle CBD $	SSS

Only two pairs were given; the third came free from the shared side. That is the move this kind of proof is built around.

Example 3: using vertical angles, then CPCTC. Segments $ AC $ and $ BD $ bisect each other at $ M $. Prove $ AB \cong CD $.

Statement	Reason
$ AM \cong MC $, $ BM \cong MD $	Definition of bisector
$ \angle AMB \cong \angle CMD $	Vertical angles
$ \triangle AMB \cong \triangle CMD $	SAS
$ AB \cong CD $	CPCTC

The vertical angle is included between the two pairs of sides, so SAS applies. CPCTC is what lets you step from "the triangles are congruent" to the specific conclusion asked for.

Practice · 15 problems

1–6 identify criteria, 7–11 supply missing facts and reasons, 12–15 reason about proofs.

Three pairs of corresponding sides are congruent. Which criterion applies?
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SSS
Two angles and the side between them are congruent. Which criterion?
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ASA
Two angles and a side not between them are congruent. Which criterion?
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AAS
Two sides and a non-included angle are congruent. What can you conclude?
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Nothing. SSA is not a valid criterion, two different triangles can satisfy it. No conclusion
All three angles are congruent. What can you conclude?
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The triangles are similar but may differ in size, so not necessarily congruent. Similar only
$ \triangle ABC \cong \triangle DEF $. Which side corresponds to $ BC $?
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Read the order: $ B \to E $ and $ C \to F $. $ EF $
Two triangles share a common side. What reason lets you call that pair congruent?
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Anything is congruent to itself. Reflexive property
Two lines cross, forming triangles on opposite sides. What angle fact is available free?
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The angles opposite each other at the intersection are equal. Vertical angles are congruent
Given $ AB \cong DE $, $ \angle A \cong \angle D $, $ AC \cong DF $, name the criterion.
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$ \angle A $ lies between sides $ AB $ and $ AC $, so the angle is included. SAS
Given $ \angle A \cong \angle D $, $ \angle B \cong \angle E $, $ AB \cong DE $, name the criterion.
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$ AB $ is the side between angles $ A $ and $ B $. ASA
Given $ \angle A \cong \angle D $, $ \angle B \cong \angle E $, $ BC \cong EF $, name the criterion.
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$ BC $ is not between the two given angles. AAS
After proving $ \triangle PQR \cong \triangle STU $, what reason justifies $ \angle Q \cong \angle T $?
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Corresponding parts of congruent triangles are congruent, used after the congruence is established. CPCTC
A student uses CPCTC as the reason for their second line, before proving the triangles congruent. What is wrong?
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CPCTC assumes the very thing being proved. It may only be used after a congruence criterion has been applied, to draw further conclusions. Circular, only valid after congruence
A student writes $ \triangle ABC \cong \triangle EFD $ when $ A $ corresponds to $ D $, $ B $ to $ E $, $ C $ to $ F $. What is wrong?
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The vertex order must reflect the correspondence. It should be $ \triangle ABC \cong \triangle DEF $. As written it claims $ A $ matches $ E $, which is false. $ \triangle ABC \cong \triangle DEF $
A student proves two triangles congruent by SSA. Draw out why that fails.
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Fix an angle and the side next to it, then swing the third side of fixed length across; it can meet the far ray in two different places, producing two non-congruent triangles with identical SSA data. Since the data does not determine the triangle, it cannot prove congruence. SSA is ambiguous

Topic 6.3 · Unit 6 · CA CCSS G-CO.12–13

Compass and straightedge constructions

A construction uses only an unmarked straightedge and a compass, no measuring. Because you cannot practice the drawing itself on a screen, this topic concentrates on the part exams actually ask about: the sequence of steps, and why each construction works.

The method

The single idea underneath everything: a compass draws all points at a fixed distance from a center, so every arc you draw is a promise that certain lengths are equal. Constructions work because those equal lengths force congruent triangles.

Copy a segment: draw a ray, set the compass to the original's length, and mark that distance from the ray's endpoint.
Perpendicular bisector of $ AB $: open the compass to more than half of $ AB $; draw arcs from $ A $ and from $ B $ on both sides; join the two crossing points. Works because each crossing point is equidistant from $ A $ and $ B $, and the set of such points is exactly the perpendicular bisector.
Angle bisector: arc from the vertex crossing both rays; from those two crossings, equal arcs meeting inside; join the vertex to that meeting point. Works because it builds two congruent triangles by SSS.
Perpendicular through a point on a line: arc from the point cutting the line either side, then bisect the segment those cuts create.
Parallel through a point: copy the angle the line makes with a transversal at the new point. Works because congruent corresponding angles force the lines parallel.

Where marks are lost: changing the compass width mid-step when the construction requires it fixed, and opening to less than half of $ AB $ for a perpendicular bisector, the arcs then never meet. If your arcs do not cross, that is the reason.

Worked examples

Example 1: why the perpendicular bisector construction works.

The two arc-crossing points, call them $ P $ and $ Q $, were each drawn at the same radius from $ A $ and from $ B $, so $ PA = PB $ and $ QA = QB $. Any point equidistant from $ A $ and $ B $ lies on the perpendicular bisector of $ AB $, so both $ P $ and $ Q $ do, and two points determine the line. The construction is a consequence of the equidistance property, not a lucky recipe.

Example 2: why the angle bisector works.

The first arc cuts the two rays at points equidistant from the vertex $ V $, say $ X $ and $ Y $, so $ VX = VY $. The next two arcs have equal radius and meet at $ Z $, so $ XZ = YZ $. Then $ \triangle VXZ \cong \triangle VYZ $ by SSS, since $ VZ $ is shared. By CPCTC $ \angle XVZ \cong \angle YVZ $, so $ VZ $ bisects the angle.

Example 3: an equilateral triangle on segment $ AB $.

Set the compass to length $ AB $. Draw an arc centered at $ A $ and another centered at $ B $; call a crossing point $ C $. Join $ AC $ and $ BC $. Both arcs had radius $ AB $, so $ AC = AB $ and $ BC = AB $, making all three sides equal. This is why the compass width must not change between the two arcs.

Practice · 15 problems

These test the sequence and the reasoning rather than the drawing.

What two tools are permitted in a classical construction?
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No measuring is allowed, so no ruler markings and no protractor. Unmarked straightedge and compass
What does every arc drawn with a compass guarantee?
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Every point on it is the same distance from the center. Equal distances from the center
In constructing a perpendicular bisector of $ AB $, how wide must the compass be?
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More than half of $ AB $, or the arcs from the two ends never meet. More than half $ AB $
Why do the two arc-crossing points lie on the perpendicular bisector?
Show the full solution

Each was drawn at equal radius from $ A $ and $ B $, so each is equidistant from both, and the set of points equidistant from $ A $ and $ B $ is exactly that bisector. They are equidistant from $ A $ and $ B $
Which congruence criterion justifies the angle bisector construction?
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Two pairs of equal radii plus the shared side $ VZ $. SSS
After proving the two triangles congruent in the angle bisector construction, what reason gives the equal angles?
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CPCTC
To construct an equilateral triangle on $ AB $, what must the compass be set to?
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The full length $ AB $, unchanged for both arcs. Length $ AB $
Why must the compass width stay fixed while drawing those two arcs?
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The equal radii are what force $ AC = BC = AB $. Changing the width breaks the guarantee and the triangle is no longer equilateral. Equal radii give equal sides
A student's perpendicular bisector arcs do not intersect. What went wrong?
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The compass was opened to less than half of $ AB $, so the two circles are too small to overlap. Compass too narrow
How do you construct a line parallel to a given line through a point not on it?
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Draw a transversal through the point, then copy the angle it makes with the original line at the new point. Copy a corresponding angle
Why does copying that angle guarantee the lines are parallel?
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If two lines cut by a transversal have congruent corresponding angles, the lines are parallel, the converse of the corresponding angles postulate. Congruent corresponding angles
How would you construct a 45° angle using only these tools?
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Construct a perpendicular to get 90°, then bisect that angle. Perpendicular, then bisect
How would you construct a 30° angle?
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Construct an equilateral triangle, which has 60° angles, then bisect one of them. Equilateral triangle, then bisect
A student uses a ruler to measure $ AB $ and marks half its length to find the midpoint. Why is that not a construction?
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Classical construction forbids measuring, the straightedge is unmarked. The midpoint must come from the perpendicular bisector construction, which produces it without any measurement. Measuring is not permitted
A student claims a construction is valid because the result "looks right" when checked with a protractor. Why is that not a justification?
Show the full solution

A construction is justified by the geometry that forces the result (equal radii producing congruent triangles) not by measurement, which is subject to drawing error and proves nothing in general. Needs a geometric argument

Topic 6.4 · Unit 6 · CA CCSS G-GPE.4–7

Coordinate geometry

This topic closes the loop on the year: the slope work from Unit 2 becomes a tool for proving geometric facts. Nearly every question reduces to computing distances and slopes, then arguing from what they show.

The method

Distance: $ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} $, the Pythagorean theorem with the horizontal and vertical gaps as legs. Leave answers in surd form unless told otherwise.
Midpoint: $ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) $, average each coordinate. Averaging, not subtracting.
Parallel means equal slopes. Perpendicular means slopes multiplying to $ -1 $, that is negative reciprocals.
To prove a shape: decide what the definition requires, then test exactly that. Parallelogram, both pairs of opposite sides parallel. Rectangle, parallelogram plus one right angle, so add a perpendicular check. Rhombus, parallelogram with all sides equal, so add distances. Square, both.
To partition a segment in ratio $ a{:}b $ from $ A $ to $ B $, move $ \frac{a}{a+b} $ of the way: add $ \frac{a}{a+b} $ of each coordinate gap to $ A $.

Where marks are lost: proving too little. Showing one pair of sides parallel does not make a parallelogram, and showing four equal sides does not make a square, a rhombus also has four equal sides. State the definition you are using, then show every condition it names.

Worked examples

Example 1: distance and midpoint. For $ A(-2, 3) $ and $ B(4, 11) $, find $ AB $ and the midpoint.

Gaps: $ \Delta x = 4 - (-2) = 6 $, $ \Delta y = 11 - 3 = 8 $. $ d = \sqrt{36 + 64} = \sqrt{100} = 10 $.

Midpoint: $ \left( \frac{-2+4}{2}, \frac{3+11}{2} \right) = (1, 7) $. Note the midpoint averages while the distance subtracts, mixing these is a common slip.

Example 2: proving a shape. Show that $ A(0,0), B(4,2), C(6,6), D(2,4) $ is a parallelogram but not a rectangle.

Slopes: $ AB = \frac{2}{4} = \frac{1}{2} $; $ DC = \frac{6-4}{6-2} = \frac{1}{2} $. Equal, so $ AB \parallel DC $. Then $ BC = \frac{6-2}{6-4} = 2 $ and $ AD = \frac{4}{2} = 2 $. Equal, so $ BC \parallel AD $. Both pairs of opposite sides are parallel, so it is a parallelogram.

For a rectangle we would need a right angle. Adjacent slopes $ \frac{1}{2} $ and 2 multiply to 1, not $ -1 $, so the sides are not perpendicular. Parallelogram, not a rectangle.

Example 3: partitioning a segment. Find the point that divides $ A(1, 2) $ to $ B(9, 10) $ in the ratio $ 3{:}1 $.

The point is $ \frac{3}{3+1} = \frac{3}{4} $ of the way along. Gaps are $ \Delta x = 8 $ and $ \Delta y = 8 $, so move $ \frac{3}{4} \times 8 = 6 $ in each.

Result: $ (1 + 6, 2 + 6) = (7, 8) $. Sanity check: it should sit closer to $ B $, which it does.

Practice · 15 problems

1–6 compute, 7–11 test relationships, 12–15 prove and diagnose.

Find the distance between $ (0,0) $ and $ (3,4) $.
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$ \sqrt{9 + 16} = \sqrt{25} = 5 $. 5
Find the distance between $ (1, 2) $ and $ (7, 10) $.
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Gaps 6 and 8: $ \sqrt{36 + 64} = 10 $. 10
Find the midpoint of $ (2, 6) $ and $ (8, 10) $.
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Average each: $ (5, 8) $. $ (5, 8) $
Find the midpoint of $ (-3, 5) $ and $ (7, -1) $.
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$ \left( \frac{-3+7}{2}, \frac{5-1}{2} \right) = (2, 2) $. $ (2, 2) $
Find the distance between $ (-2, 1) $ and $ (3, 13) $.
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Gaps 5 and 12: $ \sqrt{25 + 144} = \sqrt{169} = 13 $. 13
Find the distance between $ (1, 1) $ and $ (4, 5) $.
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Gaps 3 and 4: $ \sqrt{9+16} = 5 $. 5
Are the lines through $ (0,1),(2,5) $ and through $ (1,0),(3,4) $ parallel?
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Slopes $ \frac{4}{2} = 2 $ and $ \frac{4}{2} = 2 $. Equal. Yes
Are the lines with slopes $ \frac{3}{4} $ and $ -\frac{4}{3} $ perpendicular?
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Their product is $ -1 $. Yes
Are the lines with slopes 2 and $ \frac{1}{2} $ perpendicular?
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Product is 1, not $ -1 $. The reciprocal alone is not enough, the sign must change. No
A triangle has vertices $ (0,0), (6,0), (0,8) $. Find its perimeter.
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Legs 6 and 8 along the axes; the hypotenuse is $ \sqrt{36+64} = 10 $. Perimeter $ = 6 + 8 + 10 = 24 $. 24
Show that $ (0,0), (4,0), (4,3), (0,3) $ is a rectangle.
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Bottom and top are horizontal, slope 0; left and right are vertical, undefined slope. Both pairs of opposite sides are parallel, so it is a parallelogram, and a horizontal meeting a vertical is a right angle. Rectangle
Find the point one quarter of the way from $ (2, 3) $ to $ (10, 11) $.
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Gaps are 8 and 8; a quarter of each is 2. $ (2+2, 3+2) = (4, 5) $. $ (4, 5) $
A quadrilateral has all four sides of length 5. Is it necessarily a square?
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No, a rhombus also has four equal sides but no right angles. A perpendicularity check on adjacent sides is required. No; could be a rhombus
A student finds the midpoint of $ (2,6) $ and $ (8,10) $ as $ (6, 4) $. Find the error.
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They subtracted instead of averaging, that is the gap between the points, not the middle. The midpoint averages: $ (5, 8) $. A quick check is that the midpoint must lie between the two points in both coordinates, and 4 is not between 6 and 10. $ (5, 8) $
A student proves a quadrilateral is a parallelogram by showing one pair of opposite sides is parallel. Why is that insufficient?
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A trapezoid has exactly one pair of parallel sides and is not a parallelogram. The definition requires both pairs, so the second pair must be checked as well. Could be a trapezoid

Unit 6 mixed review · 10 problems · all topics

Unit 6 mixed review: Transformations, Congruence and Coordinates

Rigid motions, congruence criteria, constructions and coordinate proof.

Reflect $ (4, -7) $ across the $ y $-axis.
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$ (-x, y) $. $ (-4, -7) $
Rotate $ (5, 2) $ by 180° about the origin.
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$ (-x, -y) $. $ (-5, -2) $
$ (3, 1) $ maps to $ (-1, 3) $. Name the transformation.
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Matches $ (x,y) \to (-y,x) $. 90° counter-clockwise rotation about the origin
Two sides and the included angle are congruent. Which criterion applies?
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SAS
Why does SSA fail to prove congruence?
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Two different triangles can share the same two sides and non-included angle, so the data does not determine the triangle. It is ambiguous
Find the distance between $ (-1, 2) $ and $ (5, 10) $.
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Gaps 6 and 8: $ \sqrt{36 + 64} = 10 $. 10
Find the midpoint of $ (-4, 3) $ and $ (10, 9) $.
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$ \left( \frac{-4+10}{2}, \frac{3+9}{2} \right) = (3, 6) $. $ (3, 6) $
Two adjacent sides of a quadrilateral have slopes $ \frac{2}{5} $ and $ -\frac{5}{2} $. What does that show?
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Their product is $ -1 $, so the sides are perpendicular and meet at a right angle. A right angle
Which congruence criterion justifies the angle bisector construction, and why?
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SSS, two pairs of sides come from equal compass radii and the third is the shared segment from the vertex. SSS
A student shows all four sides of a quadrilateral are equal and concludes it is a square. What is missing?
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A rhombus also has four equal sides. A right angle must be shown as well, by checking that adjacent slopes are negative reciprocals. Needs a perpendicularity check

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Statement	Reason
\( AB \cong CB \)	Given
\( AD \cong CD \)	Given
\( BD \cong BD \)	Reflexive property
\( \triangle ABD \cong \triangle CBD \)	SSS

Statement	Reason
\( AM \cong MC \), \( BM \cong MD \)	Definition of bisector
\( \angle AMB \cong \angle CMD \)	Vertical angles
\( \triangle AMB \cong \triangle CMD \)	SAS
\( AB \cong CD \)	CPCTC