Pre-algebra Worksheet - Free Printable

Let's solve each equation step-by-step for the variable $ y $. We'll go through all 26 problems and explain the solution process clearly.

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1. $ 0 \times y = 0 $

Any number multiplied by 0 is 0, so this is true for any value of $ y $.
But since we're solving for $ y $, and the equation holds for all $ y $, it's an identity.
✔ Answer: All real numbers (or "infinite solutions")

> But in pre-algebra context, sometimes they expect a specific answer. Since $ 0 \times y = 0 $ is always true, any number works, but typically we say:
> ✔ Answer: Any value of $ y $ (or "no unique solution")

However, if only one answer is expected, it might be interpreted as undefined or any value — but let’s keep it as:
👉 Answer: Any real number

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2. $ 8 - 1y = 2 $

Rewrite:
$ 8 - y = 2 $
Subtract 8 from both sides:
$ -y = 2 - 8 = -6 $
Multiply both sides by -1:
$ y = 6 $

✔ Answer: $ y = 6 $

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3. $ 63 = y \times 9 $

Divide both sides by 9:
$ y = \frac{63}{9} = 7 $

✔ Answer: $ y = 7 $

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4. $ 6 - y = 7 $

Subtract 6 from both sides:
$ -y = 1 $
Multiply by -1:
$ y = -1 $

✔ Answer: $ y = -1 $

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5. $ 5 - 3y = 2 $

Subtract 5 from both sides:
$ -3y = 2 - 5 = -3 $
Divide by -3:
$ y = \frac{-3}{-3} = 1 $

✔ Answer: $ y = 1 $

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6. $ 4 = 22 - 6y $

Subtract 22 from both sides:
$ 4 - 22 = -6y $ → $ -18 = -6y $
Divide by -6:
$ y = \frac{-18}{-6} = 3 $

✔ Answer: $ y = 3 $

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7. $ 2 = y - (-1) $

Simplify right side:
$ 2 = y + 1 $
Subtract 1:
$ y = 1 $

✔ Answer: $ y = 1 $

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8. $ 6 = y \times (-2) $

So:
$ 6 = -2y $
Divide both sides by -2:
$ y = \frac{6}{-2} = -3 $

✔ Answer: $ y = -3 $

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9. $ 0 = 9 - y $

Add $ y $ to both sides:
$ y = 9 $

✔ Answer: $ y = 9 $

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10. $ 10 = 8 - y $

Subtract 8:
$ 10 - 8 = -y $ → $ 2 = -y $
Multiply by -1:
$ y = -2 $

✔ Answer: $ y = -2 $

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11. $ 0 = 21 - 7y $

Subtract 21:
$ -21 = -7y $
Divide by -7:
$ y = \frac{-21}{-7} = 3 $

✔ Answer: $ y = 3 $

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12. $ 2 = 5y - 8 $

Add 8 to both sides:
$ 2 + 8 = 5y $ → $ 10 = 5y $
Divide by 5:
$ y = 2 $

✔ Answer: $ y = 2 $

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13. $ y - 1 = 8 $

Add 1:
$ y = 9 $

✔ Answer: $ y = 9 $

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14. $ y - 4 = 3 $

Add 4:
$ y = 7 $

✔ Answer: $ y = 7 $

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15. $ 4 = y - 5 $

Add 5:
$ y = 9 $

✔ Answer: $ y = 9 $

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16. $ 5y - 6 = 4 $

Add 6:
$ 5y = 10 $
Divide by 5:
$ y = 2 $

✔ Answer: $ y = 2 $

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17. $ 1y - 1 = 5 $

So:
$ y - 1 = 5 $
Add 1:
$ y = 6 $

✔ Answer: $ y = 6 $

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18. $ 2y - 5 = 3 $

Add 5:
$ 2y = 8 $
Divide by 2:
$ y = 4 $

✔ Answer: $ y = 4 $

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19. $ 12 = y \times 6 $

So:
$ 12 = 6y $
Divide by 6:
$ y = 2 $

✔ Answer: $ y = 2 $

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20. $ 9 = 9 - (-3y) $

Simplify:
$ 9 = 9 + 3y $
Subtract 9:
$ 0 = 3y $
$ y = 0 $

✔ Answer: $ y = 0 $

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21. $ 1 = 1y - (-2) $

Simplify:
$ 1 = y + 2 $
Subtract 2:
$ y = -1 $

✔ Answer: $ y = -1 $

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22. $ 8 - y = 6 $

Subtract 8:
$ -y = -2 $
Multiply by -1:
$ y = 2 $

✔ Answer: $ y = 2 $

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23. $ 3 = 8 - y $

Subtract 8:
$ -5 = -y $
Multiply by -1:
$ y = 5 $

✔ Answer: $ y = 5 $

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24. $ y \times 9 = 27 $

So:
$ 9y = 27 $
Divide by 9:
$ y = 3 $

✔ Answer: $ y = 3 $

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25. $ y \times 0 = 0 $

Again, $ 0 \times y = 0 $ is true for all values of $ y $.
So no unique solution.

✔ Answer: Any real number (or "infinite solutions")

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26. $ 6 = 9 - y $

Subtract 9:
$ 6 - 9 = -y $ → $ -3 = -y $
Multiply by -1:
$ y = 3 $

✔ Answer: $ y = 3 $

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## ✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | Any real number |
| 2 | $ y = 6 $ |
| 3 | $ y = 7 $ |
| 4 | $ y = -1 $ |
| 5 | $ y = 1 $ |
| 6 | $ y = 3 $ |
| 7 | $ y = 1 $ |
| 8 | $ y = -3 $ |
| 9 | $ y = 9 $ |
| 10 | $ y = -2 $ |
| 11 | $ y = 3 $ |
| 12 | $ y = 2 $ |
| 13 | $ y = 9 $ |
| 14 | $ y = 7 $ |
| 15 | $ y = 9 $ |
| 16 | $ y = 2 $ |
| 17 | $ y = 6 $ |
| 18 | $ y = 4 $ |
| 19 | $ y = 2 $ |
| 20 | $ y = 0 $ |
| 21 | $ y = -1 $ |
| 22 | $ y = 2 $ |
| 23 | $ y = 5 $ |
| 24 | $ y = 3 $ |
| 25 | Any real number |
| 26 | $ y = 3 $ |

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🔍 Explanation Summary:

- We used basic algebraic operations: addition, subtraction, multiplication, division.
- Isolated the variable $ y $ on one side.
- Remembered that subtracting a negative is adding: $ -(-a) = +a $
- For equations like $ 0 \times y = 0 $, any value of $ y $ satisfies it, so no unique solution.

Let me know if you'd like this formatted as a printable worksheet!