Question

2026 algebra 2 honors test 1 - expressions, linear equations, and linear inequalities
name _________________________ date ___________ period ___
direction(s):

• make sure you carefully read each instruction before you attempt to answer a question.
• remember to show all your work to earn full or partial credit; no work or solutions with no work means no credit.
• remember that you shall not use any electronics including cell phones
• cheating in any way including conversing with others will automatically result in a zero grade.

solve each equation given below

1. $-5(x - 2) - (x + 2) = 50$
2. $3w - (7w + 12) = 2(w - 3)$
3. $8 - 3(k + 2) = 2 - 3k$
4. $-7(a - 3) = 11 - 7a$
5. $14w - 2(1 - w) = 2(5w - 1)$
6. $\frac{6}{y - 5} = \frac{18}{y + 1}$

Explanation:

Step1: Expand parentheses

$-5(x-2)-(x+2)=50$
$-5x + 10 - x - 2 = 50$

Step2: Combine like terms

$-6x + 8 = 50$

Step3: Isolate variable term

$-6x = 50 - 8$
$-6x = 42$

Step4: Solve for $x$

$x = \frac{42}{-6}$
$x = -7$

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Step1: Expand parentheses

$3w - (7w + 12) = 2(w - 3)$
$3w - 7w - 12 = 2w - 6$

Step2: Combine like terms

$-4w - 12 = 2w - 6$

Step3: Move variable terms to left

$-4w - 2w = -6 + 12$
$-6w = 6$

Step4: Solve for $w$

$w = \frac{6}{-6}$
$w = -1$

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Step1: Expand parentheses

$8 - 3(k + 2) = 2 - 3k$
$8 - 3k - 6 = 2 - 3k$

Step2: Combine like terms

$2 - 3k = 2 - 3k$

Step3: Simplify equation

$0 = 0$
This is always true, so all real numbers are solutions.

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Step1: Expand parentheses

$-7(a - 3) = 11 - 7a$
$-7a + 21 = 11 - 7a$

Step2: Move variable terms to left

$-7a + 7a = 11 - 21$
$0 = -10$
This is never true, so there is no solution.

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Step1: Expand parentheses

$14w - 2(1 - w) = 2(5w - 1)$
$14w - 2 + 2w = 10w - 2$

Step2: Combine like terms

$16w - 2 = 10w - 2$

Step3: Isolate variable term

$16w - 10w = -2 + 2$
$6w = 0$

Step4: Solve for $w$

$w = \frac{0}{6}$
$w = 0$

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Step1: Cross-multiply to eliminate fractions

$\frac{6}{y-5} = \frac{18}{y+1}$
$6(y + 1) = 18(y - 5)$

Step2: Expand both sides

$6y + 6 = 18y - 90$

Step3: Isolate variable term

$6 + 90 = 18y - 6y$
$96 = 12y$

Step4: Solve for $y$

$y = \frac{96}{12}$
$y = 8$
(Verify: $y=8$ does not make denominators zero, so it is valid)

Answer:

1. $x=-7$
2. $w=-1$
3. All real numbers
4. No solution
5. $w=0$
6. $y=8$