Worksheet with algebraic equations requiring solving for specified variables.

Let's solve each equation step by step, isolating the requested variable. I'll go through each one systematically.

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1. $ 2x + a = b $ — Solve for $ x $

Subtract $ a $ from both sides:
$$
2x = b - a
$$
Divide both sides by 2:
$$
x = \frac{b - a}{2}
$$

✔ Answer: $ x = \frac{b - a}{2} $

---

2. $ ax + b = c + d $ — Solve for $ a $

Subtract $ b $ from both sides:
$$
ax = c + d - b
$$
Divide both sides by $ x $ (assuming $ x \neq 0 $):
$$
a = \frac{c + d - b}{x}
$$

✔ Answer: $ a = \frac{c + d - b}{x} $

---

3. $ S = 4x $ — Solve for $ x $

Divide both sides by 4:
$$
x = \frac{S}{4}
$$

✔ Answer: $ x = \frac{S}{4} $

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4. $ 36 = byx $ — Solve for $ y $

Divide both sides by $ bx $ (assuming $ b \neq 0, x \neq 0 $):
$$
y = \frac{36}{bx}
$$

✔ Answer: $ y = \frac{36}{bx} $

---

5. $ x + sy = t $ — Solve for $ y $

Subtract $ x $ from both sides:
$$
sy = t - x
$$
Divide by $ s $ (assuming $ s \neq 0 $):
$$
y = \frac{t - x}{s}
$$

✔ Answer: $ y = \frac{t - x}{s} $

---

6. $ cm - d = 4 $ — Solve for $ m $

Add $ d $ to both sides:
$$
cm = 4 + d
$$
Divide by $ c $ (assuming $ c \neq 0 $):
$$
m = \frac{4 + d}{c}
$$

✔ Answer: $ m = \frac{4 + d}{c} $

---

7. $ Bx - 5 = c $ — Solve for $ y $

Wait: The equation is in terms of $ x $, but we're asked to solve for $ y $. There is no $ y $ in this equation.

✘ Error: This equation does not contain $ y $. Cannot solve for $ y $.

But perhaps it's a typo? If the intention was to solve for $ x $, then:

Add 5 to both sides:
$$
Bx = c + 5
$$
Divide by $ B $:
$$
x = \frac{c + 5}{B}
$$

But since the instruction says "solve for $ y $" and there is no $ y $, we must assume either a typo or that $ y $ is unrelated.

⚠️ Conclusion: Cannot solve for $ y $ — it's not in the equation.

However, if the problem meant to say "solve for $ x $" instead, then:
✔ $ x = \frac{c + 5}{B} $

But based on what's written, we can't proceed.

Let’s assume it's a typo and proceed with solving for $ x $, or flag it.

But let's double-check: Is it possible $ y $ is a typo for $ x $? Possibly.

But since the original says “solve for $ y $”, and $ y $ doesn't appear, this is invalid.

So unless there’s a mistake in transcription, we skip or note the issue.

But looking at the pattern, maybe it's supposed to be $ By - 5 = c $? That would make sense.

But as written: $ Bx - 5 = c $, solve for $ y $ → No solution possible.

We’ll leave it as invalid unless corrected.

But for now, we'll assume it's a typo and they meant to solve for $ x $. So:

✔ $ x = \frac{c + 5}{B} $

(But technically, per instructions, this one is flawed.)

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8. $ m = 2(x + n) $ — Solve for $ n $

Divide both sides by 2:
$$
\frac{m}{2} = x + n
$$
Subtract $ x $:
$$
n = \frac{m}{2} - x
$$

✔ Answer: $ n = \frac{m}{2} - x $

---

9. $ 4p - 5c = 3c $ — Solve for $ p $

Add $ 5c $ to both sides:
$$
4p = 3c + 5c = 8c
$$
Divide by 4:
$$
p = \frac{8c}{4} = 2c
$$

✔ Answer: $ p = 2c $

---

10. $ abx - d = 5d $ — Solve for $ b $

Add $ d $ to both sides:
$$
abx = 5d + d = 6d
$$
Divide both sides by $ ax $ (assuming $ a \neq 0, x \neq 0 $):
$$
b = \frac{6d}{ax}
$$

✔ Answer: $ b = \frac{6d}{ax} $

---

11. $ 3z - s = r $ — Solve for $ z $

Add $ s $ to both sides:
$$
3z = r + s
$$
Divide by 3:
$$
z = \frac{r + s}{3}
$$

✔ Answer: $ z = \frac{r + s}{3} $

---

12. $ \frac{1}{4}y + bx = c $ — Solve for $ y $

Subtract $ bx $ from both sides:
$$
\frac{1}{4}y = c - bx
$$
Multiply both sides by 4:
$$
y = 4(c - bx) = 4c - 4bx
$$

✔ Answer: $ y = 4c - 4bx $

---

13. $ m = Dv $ — Solve for $ D $

Divide both sides by $ v $ (assuming $ v \neq 0 $):
$$
D = \frac{m}{v}
$$

✔ Answer: $ D = \frac{m}{v} $

---

14. $ px + 9j = 6c $ — Solve for $ p $

Subtract $ 9j $ from both sides:
$$
px = 6c - 9j
$$
Divide by $ x $ (assuming $ x \neq 0 $):
$$
p = \frac{6c - 9j}{x}
$$

✔ Answer: $ p = \frac{6c - 9j}{x} $

---

15. $ 4ad + 4d = 8q $ — Solve for $ a $

Factor out $ 4d $ on the left:
$$
4d(a + 1) = 8q
$$
Divide both sides by $ 4d $ (assuming $ d \neq 0 $):
$$
a + 1 = \frac{8q}{4d} = \frac{2q}{d}
$$
Subtract 1:
$$
a = \frac{2q}{d} - 1
$$

✔ Answer: $ a = \frac{2q}{d} - 1 $

---

16. $ hy + 7z = 14 $ — Solve for $ y $

Subtract $ 7z $ from both sides:
$$
hy = 14 - 7z
$$
Divide by $ h $ (assuming $ h \neq 0 $):
$$
y = \frac{14 - 7z}{h}
$$

✔ Answer: $ y = \frac{14 - 7z}{h} $

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## ✔ Final Answers Summary:

| Equation | Solve for | Answer |
|--------|----------|--------|
| $ 2x + a = b $ | $ x $ | $ x = \frac{b - a}{2} $ |
| $ ax + b = c + d $ | $ a $ | $ a = \frac{c + d - b}{x} $ |
| $ S = 4x $ | $ x $ | $ x = \frac{S}{4} $ |
| $ 36 = byx $ | $ y $ | $ y = \frac{36}{bx} $ |
| $ x + sy = t $ | $ y $ | $ y = \frac{t - x}{s} $ |
| $ cm - d = 4 $ | $ m $ | $ m = \frac{4 + d}{c} $ |
| $ Bx - 5 = c $ | $ y $ | ✘ Invalid — no $ y $ in equation |
| $ m = 2(x + n) $ | $ n $ | $ n = \frac{m}{2} - x $ |
| $ 4p - 5c = 3c $ | $ p $ | $ p = 2c $ |
| $ abx - d = 5d $ | $ b $ | $ b = \frac{6d}{ax} $ |
| $ 3z - s = r $ | $ z $ | $ z = \frac{r + s}{3} $ |
| $ \frac{1}{4}y + bx = c $ | $ y $ | $ y = 4c - 4bx $ |
| $ m = Dv $ | $ D $ | $ D = \frac{m}{v} $ |
| $ px + 9j = 6c $ | $ p $ | $ p = \frac{6c - 9j}{x} $ |
| $ 4ad + 4d = 8q $ | $ a $ | $ a = \frac{2q}{d} - 1 $ |
| $ hy + 7z = 14 $ | $ y $ | $ y = \frac{14 - 7z}{h} $ |

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⚠️ Note: For equation #7 ($ Bx - 5 = c $), solving for $ y $ is not possible because $ y $ does not appear in the equation. It may be a typo — likely intended to be solve for $ x $, in which case $ x = \frac{c + 5}{B} $.

Let me know if you'd like a cleaned-up version assuming corrections.