Question

solve the equation. then check your solution.

1. $\frac{d}{10}=12$
2. $-5=\frac{n}{4}$
3. $-15=\frac{x}{-8}$
4. $\frac{t}{28}=9$
5. $\frac{x}{-6}=13$
6. $-12=\frac{a}{-3}$
7. $\frac{b}{3}=8.2$
8. $\frac{x}{5.3}=0.7$
9. $\frac{p}{2.3}=4$
10. $\frac{y}{50}=-21$
11. $\frac{1}{2}=\frac{d}{4}$
12. $\frac{3}{4}=\frac{a}{8}$

Explanation:

Step1: Isolate variable via cross-multiply

For $\frac{d}{10}=12$:
$d = 12 \times 10$

Step2: Calculate the product

$d = 120$

Step3: Verify solution

Substitute $d=120$: $\frac{120}{10}=12$, which is true.

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Step1: Isolate variable via cross-multiply

For $-5=\frac{n}{4}$:
$n = -5 \times 4$

Step2: Calculate the product

$n = -20$

Step3: Verify solution

Substitute $n=-20$: $\frac{-20}{4}=-5$, which is true.

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Step1: Isolate variable via cross-multiply

For $-15=\frac{x}{-8}$:
$x = -15 \times (-8)$

Step2: Calculate the product

$x = 120$

Step3: Verify solution

Substitute $x=120$: $\frac{120}{-8}=-15$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{t}{28}=9$:
$t = 9 \times 28$

Step2: Calculate the product

$t = 252$

Step3: Verify solution

Substitute $t=252$: $\frac{252}{28}=9$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{x}{-6}=13$:
$x = 13 \times (-6)$

Step2: Calculate the product

$x = -78$

Step3: Verify solution

Substitute $x=-78$: $\frac{-78}{-6}=13$, which is true.

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Step1: Isolate variable via cross-multiply

For $-12=\frac{a}{-3}$:
$a = -12 \times (-3)$

Step2: Calculate the product

$a = 36$

Step3: Verify solution

Substitute $a=36$: $\frac{36}{-3}=-12$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{b}{3}=8.2$:
$b = 8.2 \times 3$

Step2: Calculate the product

$b = 24.6$

Step3: Verify solution

Substitute $b=24.6$: $\frac{24.6}{3}=8.2$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{x}{5.3}=0.7$:
$x = 0.7 \times 5.3$

Step2: Calculate the product

$x = 3.71$

Step3: Verify solution

Substitute $x=3.71$: $\frac{3.71}{5.3}=0.7$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{p}{2.3}=4$:
$p = 4 \times 2.3$

Step2: Calculate the product

$p = 9.2$

Step3: Verify solution

Substitute $p=9.2$: $\frac{9.2}{2.3}=4$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{y}{50}=-21$:
$y = -21 \times 50$

Step2: Calculate the product

$y = -1050$

Step3: Verify solution

Substitute $y=-1050$: $\frac{-1050}{50}=-21$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{1}{2}=\frac{d}{4}$:
$d = \frac{1}{2} \times 4$

Step2: Calculate the product

$d = 2$

Step3: Verify solution

Substitute $d=2$: $\frac{2}{4}=\frac{1}{2}$, which is true.

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Step1: Isolate variable via cross-multiply

For $\frac{3}{4}=\frac{a}{8}$:
$a = \frac{3}{4} \times 8$

Step2: Calculate the product

$a = 6$

Step3: Verify solution

Substitute $a=6$: $\frac{6}{8}=\frac{3}{4}$, which is true.

Answer:

1. $d=120$
2. $n=-20$
3. $x=120$
4. $t=252$
5. $x=-78$
6. $a=36$
7. $b=24.6$
8. $x=3.71$
9. $p=9.2$
10. $y=-1050$
11. $d=2$
12. $a=6$