Rearranging Equations worksheet featuring 12 algebra problems to solve for x.

Let's solve each equation by rearranging to make the specified variable the subject. I'll go through each problem step-by-step.

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1. $ 4x = c + a $

We want to make $ x $ the subject.

- Divide both sides by 4:
$$
x = \frac{c + a}{4}
$$

✔ Answer: $ x = \frac{c + a}{4} $

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2. $ 3 = \frac{y}{z} + 2 $

Make $ y $ the subject.

- Subtract 2 from both sides:
$$
3 - 2 = \frac{y}{z} \Rightarrow 1 = \frac{y}{z}
$$
- Multiply both sides by $ z $:
$$
y = z
$$

✔ Answer: $ y = z $

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3. $ \frac{a}{b} = 8 $

Make $ a $ the subject.

- Multiply both sides by $ b $:
$$
a = 8b
$$

✔ Answer: $ a = 8b $

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4. $ \frac{1}{2}x + a = 6 $

Make $ x $ the subject.

- Subtract $ a $ from both sides:
$$
\frac{1}{2}x = 6 - a
$$
- Multiply both sides by 2:
$$
x = 2(6 - a) = 12 - 2a
$$

✔ Answer: $ x = 12 - 2a $

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5. $ \frac{10}{d} = c $

Make $ d $ the subject.

- Multiply both sides by $ d $:
$$
10 = c \cdot d
$$
- Divide both sides by $ c $:
$$
d = \frac{10}{c}
$$

✔ Answer: $ d = \frac{10}{c} $

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6. $ -2(x + 1) = 10x $

Make $ x $ the subject.

- Expand the left side:
$$
-2x - 2 = 10x
$$
- Add $ 2x $ to both sides:
$$
-2 = 12x
$$
- Divide by 12:
$$
x = -\frac{2}{12} = -\frac{1}{6}
$$

But wait — this is solving for $ x $, not rearranging. The question says "make $ x $ the subject", but in this case, it's already an equation with only $ x $. So we solve:

✔ Answer: $ x = -\frac{1}{6} $

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7. $ 1 - 10x = 2x + 4 $

Make $ x $ the subject.

- Move all $ x $ terms to one side and constants to the other:
$$
1 - 4 = 2x + 10x \Rightarrow -3 = 12x
$$
- Divide by 12:
$$
x = -\frac{3}{12} = -\frac{1}{4}
$$

✔ Answer: $ x = -\frac{1}{4} $

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8. $ 8x + 2 = 3x $

Make $ x $ the subject.

- Subtract $ 3x $ from both sides:
$$
5x + 2 = 0
$$
- Subtract 2:
$$
5x = -2
$$
- Divide by 5:
$$
x = -\frac{2}{5}
$$

✔ Answer: $ x = -\frac{2}{5} $

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9. $ -x + y = \frac{1}{2} $

Make $ x $ the subject.

- Subtract $ y $ from both sides:
$$
-x = \frac{1}{2} - y
$$
- Multiply both sides by $ -1 $:
$$
x = y - \frac{1}{2}
$$

✔ Answer: $ x = y - \frac{1}{2} $

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10. $ 5(x - 2) = 4 + q $

Make $ x $ the subject.

- Expand the left side:
$$
5x - 10 = 4 + q
$$
- Add 10 to both sides:
$$
5x = 14 + q
$$
- Divide by 5:
$$
x = \frac{14 + q}{5}
$$

✔ Answer: $ x = \frac{14 + q}{5} $

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11. $ 10x - x = x + 0 $

Simplify both sides:

- Left: $ 9x $
- Right: $ x $
- So: $ 9x = x $
- Subtract $ x $: $ 8x = 0 $
- $ x = 0 $

✔ Answer: $ x = 0 $

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12. $ 4b - 2p = -27q $

Make $ p $ the subject.

- Subtract $ 4b $ from both sides:
$$
-2p = -27q - 4b
$$
- Divide both sides by $ -2 $:
$$
p = \frac{27q + 4b}{2}
$$

✔ Answer: $ p = \frac{27q + 4b}{2} $

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✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | $ x = \frac{c + a}{4} $ |
| 2 | $ y = z $ |
| 3 | $ a = 8b $ |
| 4 | $ x = 12 - 2a $ |
| 5 | $ d = \frac{10}{c} $ |
| 6 | $ x = -\frac{1}{6} $ |
| 7 | $ x = -\frac{1}{4} $ |
| 8 | $ x = -\frac{2}{5} $ |
| 9 | $ x = y - \frac{1}{2} $ |
| 10 | $ x = \frac{14 + q}{5} $ |
| 11 | $ x = 0 $ |
| 12 | $ p = \frac{27q + 4b}{2} $ |

Let me know if you'd like these written in a formatted worksheet or need explanations for any specific steps!