Pre-Algebra Equations: Solve for the variable using inverse operations.

Let's solve each of these pre-algebra equations step by step using inverse operations. The goal is to isolate the variable $ y $ in each equation.

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1. $ 13^2 - 9xy = 70 $

Wait — this equation has both $ x $ and $ y $, but only one equation. Since it's asking to "solve for the variable" (singular), and there are two variables, we need to check if there's a typo or if we're supposed to treat $ x $ as known.

But looking at the other problems, they all seem to involve only $ y $ as the unknown. So likely, this is a typo: perhaps it should be $ 13^2 - 9y = 70 $, not $ 9xy $. Let's assume that.

So:
$$
13^2 - 9y = 70
$$

Step 1: Compute $ 13^2 $
$$
169 - 9y = 70
$$

Step 2: Subtract 169 from both sides
$$
-9y = 70 - 169 = -99
$$

Step 3: Divide both sides by -9
$$
y = \frac{-99}{-9} = 11
$$

✔ Answer: $ y = 11 $

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2. $ 2^2 + y \times 1 = 15 $

Simplify:
$$
4 + y = 15
$$

Subtract 4:
$$
y = 15 - 4 = 11
$$

✔ Answer: $ y = 11 $

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3. $ 8x + y = 27 $

Again, two variables. But likely meant to be $ 8y + y = 27 $? That would make sense.

Assuming it's $ 8y + y = 27 $

Then:
$$
9y = 27 \Rightarrow y = 3
$$

✔ Answer: $ y = 3 $

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4. $ 6x + 12 = 48 $

This one has only $ x $, but again, we're solving for $ y $? Wait — no variable $ y $. Likely typo.

Possibly meant to be $ 6y + 12 = 48 $

Assume:
$$
6y + 12 = 48
$$

Subtract 12:
$$
6y = 36
\Rightarrow y = 6
$$

✔ Answer: $ y = 6 $

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5. $ 13x - 7 = 6 $

Only $ x $ here. Probably meant $ 13y - 7 = 6 $

So:
$$
13y - 7 = 6
$$

Add 7:
$$
13y = 13
\Rightarrow y = 1
$$

✔ Answer: $ y = 1 $

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6. $ 3x + y = 40 $

Two variables. Maybe $ 3y + y = 40 $? Then:

$$
4y = 40 \Rightarrow y = 10
$$

✔ Answer: $ y = 10 $

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7. $ 9^2 + y \times 5 = 136 $

Compute $ 9^2 = 81 $

So:
$$
81 + 5y = 136
$$

Subtract 81:
$$
5y = 55
\Rightarrow y = 11
$$

✔ Answer: $ y = 11 $

---

8. $ 13x + y = 84 $

Again, two variables. Possibly $ 13y + y = 84 $? Then:

$$
14y = 84 \Rightarrow y = 6
$$

✔ Answer: $ y = 6 $

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9. $ 4x - y = 24 $

Likely meant $ 4y - y = 24 $? Then:

$$
3y = 24 \Rightarrow y = 8
$$

✔ Answer: $ y = 8 $

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10. $ 8x + 3 = 99 $

Only $ x $. Probably meant $ 8y + 3 = 99 $

So:
$$
8y + 3 = 99
\Rightarrow 8y = 96
\Rightarrow y = 12
$$

✔ Answer: $ y = 12 $

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✔ Final Answers (assuming corrected versions):

| Problem | Corrected Equation | Answer |
|--------|----------------------------|--------|
| 1 | $ 13^2 - 9y = 70 $ | $ y = 11 $ |
| 2 | $ 2^2 + y = 15 $ | $ y = 11 $ |
| 3 | $ 8y + y = 27 $ | $ y = 3 $ |
| 4 | $ 6y + 12 = 48 $ | $ y = 6 $ |
| 5 | $ 13y - 7 = 6 $ | $ y = 1 $ |
| 6 | $ 3y + y = 40 $ | $ y = 10 $ |
| 7 | $ 9^2 + 5y = 136 $ | $ y = 11 $ |
| 8 | $ 13y + y = 84 $ | $ y = 6 $ |
| 9 | $ 4y - y = 24 $ | $ y = 8 $ |
| 10 | $ 8y + 3 = 99 $ | $ y = 12 $ |

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📝 Notes:

- Many equations had $ x $ instead of $ y $, which suggests typos.
- Assuming the intent was to solve for $ y $, I replaced $ x $ with $ y $ where needed.
- All solutions used inverse operations: add/subtract, multiply/divide to isolate $ y $.

If you have the original image and want to confirm the exact equations, feel free to clarify!