Question

bell-ringer 01/15/2026 solve each equation 1) $9(-3t - 9) + 7 = -2(18t - 17)$ 2) $6(10n - 13) = -6(-7n - 11) + 19n$ 20 minutes 3) $8x - 8 = 12x - 8$ 7) $3x - 5 = 3x + 10$ 4) $3(12x - 5) = 36x - 15$ 5) $\frac{9}{k + 2} = \frac{3}{k - 2}$ 6) $\frac{z + 2}{10} = \frac{z + 4}{5}$

Explanation:

Step1: Expand both sides

$9(-3q-9)+7 = -2(18q-17)$
$-27q - 81 + 7 = -36q + 34$

Step2: Simplify constants

$-27q - 74 = -36q + 34$

Step3: Isolate q terms

$-27q + 36q = 34 + 74$
$9q = 108$

Step4: Solve for q

$q = \frac{108}{9} = 12$

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Step1: Expand both sides

$6(10n-13) = -6(-7n-11)+19n$
$60n - 78 = 42n + 66 + 19n$

Step2: Combine like terms

$60n - 78 = 61n + 66$

Step3: Isolate n terms

$60n - 61n = 66 + 78$
$-n = 144$

Step4: Solve for n

$n = -144$

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Step1: Isolate x terms

$8x - 8 = 12x - 8$
$8x - 12x = -8 + 8$

Step2: Simplify

$-4x = 0$

Step3: Solve for x

$x = 0$

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Step1: Expand left side

$3(12x-5) = 36x-15$
$36x - 15 = 36x - 15$

Step2: Simplify

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

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Step1: Cross-multiply

$\frac{9}{k+2} = \frac{3}{k-2}$
$9(k-2) = 3(k+2)$

Step2: Expand both sides

$9k - 18 = 3k + 6$

Step3: Isolate k terms

$9k - 3k = 6 + 18$
$6k = 24$

Step4: Solve for k

$k = \frac{24}{6} = 4$

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Step1: Cross-multiply

$\frac{z+2}{10} = \frac{z+4}{5}$
$5(z+2) = 10(z+4)$

Step2: Expand both sides

$5z + 10 = 10z + 40$

Step3: Isolate z terms

$5z - 10z = 40 - 10$
$-5z = 30$

Step4: Solve for z

$z = \frac{30}{-5} = -6$

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Step1: Isolate x terms

$3x - 5 = 3x + 10$
$3x - 3x = 10 + 5$

Step2: Simplify

$0 = 15$
This is never true, so there is no solution.

Answer:

1. $q=12$
2. $n=-144$
3. $x=0$
4. All real numbers
5. $k=4$
6. $z=-6$
7. No solution