Unit 2 · Checkpoint quiz

Unit 2 checkpoint: Linear equations & inequalities

12 questions · about 25-30 minutes · no time limit

Answer every question, then press Grade my answers at the bottom. You'll see what you got right, a worked solution for each question, and links to the lessons worth revisiting. Pencil and paper are encouraged.

This is practice, not a test of you. A low score just shows where to look next. The last two questions review Unit 1.

Solve $7x - 4 = 3x + 12$.

Subtract $3x$ from both sides: $4x - 4 = 12$.

Add 4: $4x = 16$. Divide by 4: $x = 4$.

Check: $7(4) - 4 = 24$ and $3(4) + 12 = 24$. Answer: $x = 4$.

Which numbers are solutions of $3(x + 2) - x = 2(x + 3)$?

Every real number works
No real number works
Only $x = 0$ works
Only $x = 6$ works

Left side: $3x + 6 - x = 2x + 6$. Right side: $2x + 6$.

Subtract $2x$ from both sides: $6 = 6$. That is always true, and $x$ is gone.

This is an identity: every real number is a solution. Answer: a.

Solve $\dfrac{x}{3} + \dfrac{x - 1}{4} = 5$.

Multiply every term by the LCD, 12: $4x + 3(x - 1) = 60$.

$4x + 3x - 3 = 60$, so $7x = 63$ and $x = 9$.

Check: $\dfrac{9}{3} + \dfrac{8}{4} = 3 + 2 = 5$. Answer: $x = 9$.

Solve $0.05x + 0.1(20 - x) = 1.6$.

Multiply every term by 100 to clear the decimals: $5x + 10(20 - x) = 160$.

$5x + 200 - 10x = 160$, so $-5x = -40$ and $x = 8$.

Check: $0.05(8) + 0.1(12) = 0.4 + 1.2 = 1.6$. Answer: $x = 8$.

Solve $3x - 2y = 8$ for $y$.

Subtract $3x$ from both sides: $-2y = -3x + 8$.

Divide every term by $-2$: $y = \dfrac{-3x}{-2} + \dfrac{8}{-2}$.

Answer: $y = \tfrac{3}{2}x - 4$ (same as $y = \dfrac{3x - 8}{2}$).

A jacket is on sale for 15% off. The sale price is $68. What was the original price, in dollars?

Let $p$ = the original price. After 15% off, you pay 85% of it: $0.85p = 68$.

Divide: $p = \dfrac{68}{0.85} = 80$.

Check: 15% of $80 is $12, and $80 − $12 = $68. Answer: $80.

Solve $-2x + 5 \gt 11$. Give the answer in interval notation.

Subtract 5: $-2x \gt 6$.

Divide by $-2$ and flip the sign (dividing by a negative): $x \lt -3$.

Answer: $(-\infty, -3)$.

Solve $x + 3 \lt 1$ or $2x - 1 \ge 5$. Give the answer in interval notation.

First part: $x + 3 \lt 1$ gives $x \lt -2$.

Second part: $2x - 1 \ge 5$ gives $2x \ge 6$, so $x \ge 3$.

"Or" means either piece works, so join them with a union. Answer: $(-\infty, -2) \cup [3, \infty)$.

Solve $2|x - 4| + 3 = 11$. List every solution.

Isolate the absolute value first: $2|x - 4| = 8$, so $|x - 4| = 4$.

Two cases: $x - 4 = 4$ gives $x = 8$; $x - 4 = -4$ gives $x = 0$.

Answer: $x = 8$ or $x = 0$.

Solve $|2x - 1| \gt 7$. Give the answer in interval notation.

"Greater than" means the inside is far from 0 in either direction: $2x - 1 \gt 7$ or $2x - 1 \lt -7$.

$2x \gt 8$ gives $x \gt 4$. $2x \lt -6$ gives $x \lt -3$.

Answer: $(-\infty, -3) \cup (4, \infty)$.

Subtract: $\dfrac{5}{6} - \dfrac{3}{4}$.

The LCD of 6 and 4 is 12.

$\dfrac{5}{6} = \dfrac{10}{12}$ and $\dfrac{3}{4} = \dfrac{9}{12}$.

$\dfrac{10}{12} - \dfrac{9}{12} = \dfrac{1}{12}$. Answer: $\dfrac{1}{12}$.

Evaluate $x^2 - 3xy$ when $x = -2$ and $y = 5$.

Substitute with parentheses: $(-2)^2 - 3(-2)(5)$.

$(-2)^2 = 4$ and $3(-2)(5) = -30$.

$4 - (-30) = 34$. Answer: $34$.

Want to look something up first? Equations and inequalities cheat sheet · Glossary

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