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

Let's solve each of these pre-algebra equations step by step using inverse operations. We'll isolate the variable on one side of the equation.

---

1. $ y^2 - (1 \times 10) = 39 $

Simplify:
$$
y^2 - 10 = 39
$$

Add 10 to both sides:
$$
y^2 = 49
$$

Take the square root of both sides:
$$
y = \pm 7
$$

✔ Answer: $ y = 7 $ or $ y = -7 $

---

2. $ 4 + 12y = 136 $

Subtract 4 from both sides:
$$
12y = 132
$$

Divide both sides by 12:
$$
y = 11
$$

✔ Answer: $ y = 11 $

---

3. $ 12 + 2y = 24 $

Subtract 12 from both sides:
$$
2y = 12
$$

Divide by 2:
$$
y = 6
$$

✔ Answer: $ y = 6 $

---

4. $ 4 + y \times 10 = 24 $

Rewrite: $ 4 + 10y = 24 $

Subtract 4:
$$
10y = 20
$$

Divide by 10:
$$
y = 2
$$

✔ Answer: $ y = 2 $

---

5. $ 7xy - y = 42 $

Factor out $ y $:
$$
y(7x - 1) = 42
$$

But wait — this equation has two variables: $ x $ and $ y $. Since only one equation is given, we can't solve for both unless there's more info.

Wait — looking back: Is this a typo? It says "solve for the variable" — but it has two variables.

But perhaps the problem meant $ 7y - y = 42 $? That would make sense.

Alternatively, if it's really $ 7xy - y = 42 $, then:

$$
y(7x - 1) = 42
\Rightarrow y = \frac{42}{7x - 1}
$$

But that's not solvable uniquely without knowing $ x $.

However, in context of pre-algebra and the rest being single-variable equations, it’s likely a typo.

Let’s assume it's:
$ 7y - y = 42 $ → $ 6y = 42 $ → $ y = 7 $

But if it's truly $ 7xy - y = 42 $, then we need more information.

Wait — let’s check the original image again: It says "7xy - y = 42", which implies two variables.

But since all others are single-variable, maybe it's supposed to be 7y - y = 42?

Alternatively, could it be $ 7x - y = 42 $? But no — it's written as 7xy.

Another possibility: Maybe it's 7 × y - y = 42 — i.e., $ 7y - y = 42 $

That makes sense.

So assuming typo, and it's:
$$
7y - y = 42
\Rightarrow 6y = 42
\Rightarrow y = 7
$$

✔ Answer: $ y = 7 $

---

6. $ y^2 - (5 \times 9) = -20 $

Simplify:
$$
y^2 - 45 = -20
$$

Add 45 to both sides:
$$
y^2 = 25
$$

Take square root:
$$
y = \pm 5
$$

✔ Answer: $ y = 5 $ or $ y = -5 $

---

7. $ 8 + y \times 2 = 12 $

Rewrite: $ 8 + 2y = 12 $

Subtract 8:
$$
2y = 4
$$

Divide by 2:
$$
y = 2
$$

✔ Answer: $ y = 2 $

---

8. $ 1 \times y - 6 = 4 $

Simplify: $ y - 6 = 4 $

Add 6:
$$
y = 10
$$

✔ Answer: $ y = 10 $

---

9. $ y^2 - (1 \times 2) = 62 $

Simplify:
$$
y^2 - 2 = 62
$$

Add 2:
$$
y^2 = 64
$$

Take square root:
$$
y = \pm 8
$$

✔ Answer: $ y = 8 $ or $ y = -8 $

---

10. $ 6 + 12y = 66 $

Subtract 6:
$$
12y = 60
$$

Divide by 12:
$$
y = 5
$$

✔ Answer: $ y = 5 $

---

## ✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1 | $ y = \pm 7 $ |
| 2 | $ y = 11 $ |
| 3 | $ y = 6 $ |
| 4 | $ y = 2 $ |
| 5 | $ y = 7 $ *(assuming typo: 7y - y = 42)* |
| 6 | $ y = \pm 5 $ |
| 7 | $ y = 2 $ |
| 8 | $ y = 10 $ |
| 9 | $ y = \pm 8 $ |
| 10 | $ y = 5 $ |

---

⚠️ Note on Problem #5: If the original equation is truly $ 7xy - y = 42 $, then it cannot be solved for a unique value of $ y $ without knowing $ x $. But given the context, it's almost certainly a typo and should be $ 7y - y = 42 $, so $ y = 7 $.

Let me know if you'd like a scanned version of the work shown step-by-step!