LITERAL EQUATIONS WORKSHEET Solve for the indicated ... - Free Printable

Let's solve each of these literal equations for the indicated variable. A literal equation is an equation that involves multiple variables, and we are solving for one specific variable in terms of the others.

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1) $ P = IRT $ (solve for $ T $)

Divide both sides by $ IR $:
$$
T = \frac{P}{IR}
$$

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2) $ A = 2(L + W) $ (solve for $ W $)

Divide both sides by 2:
$$
\frac{A}{2} = L + W
$$
Subtract $ L $:
$$
W = \frac{A}{2} - L
$$

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3) $ y = 5x - 6 $ (solve for $ x $)

Add 6 to both sides:
$$
y + 6 = 5x
$$
Divide by 5:
$$
x = \frac{y + 6}{5}
$$

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4) $ 2x - 3y = 8 $ (solve for $ y $)

Subtract $ 2x $:
$$
-3y = -2x + 8
$$
Divide by $-3$:
$$
y = \frac{2x - 8}{3}
$$

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5) $ \frac{x + y}{3} = 5 $ (solve for $ x $)

Multiply both sides by 3:
$$
x + y = 15
$$
Subtract $ y $:
$$
x = 15 - y
$$

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6) $ y = mx + b $ (solve for $ b $)

Subtract $ mx $:
$$
b = y - mx
$$

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7) $ ax + by = c $ (solve for $ y $)

Subtract $ ax $:
$$
by = c - ax
$$
Divide by $ b $:
$$
y = \frac{c - ax}{b}
$$

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8) $ A = 1.2h(b + c) $ (solve for $ b $)

Divide both sides by $ 1.2h $:
$$
\frac{A}{1.2h} = b + c
$$
Subtract $ c $:
$$
b = \frac{A}{1.2h} - c
$$

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9) $ V = LWH $ (solve for $ L $)

Divide both sides by $ WH $:
$$
L = \frac{V}{WH}
$$

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10) $ A = 4\pi r^2 $ (solve for $ r^2 $)

Divide by $ 4\pi $:
$$
r^2 = \frac{A}{4\pi}
$$

> Note: If you want $ r $, take square root: $ r = \sqrt{\frac{A}{4\pi}} $, but since it asks for $ r^2 $, this is sufficient.

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11) $ V = \pi r^2 h $ (solve for $ h $)

Divide by $ \pi r^2 $:
$$
h = \frac{V}{\pi r^2}
$$

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12) $ 7x - y = 14 $ (solve for $ x $)

Add $ y $:
$$
7x = y + 14
$$
Divide by 7:
$$
x = \frac{y + 14}{7}
$$

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13) $ A = \frac{x + y}{2} $ (solve for $ y $)

Multiply both sides by 2:
$$
2A = x + y
$$
Subtract $ x $:
$$
y = 2A - x
$$

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14) $ R = \frac{E}{I} $ (solve for $ I $)

Multiply both sides by $ I $:
$$
RI = E
$$
Divide by $ R $:
$$
I = \frac{E}{R}
$$

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15) $ x = \frac{yz}{6} $ (solve for $ z $)

Multiply both sides by 6:
$$
6x = yz
$$
Divide by $ y $:
$$
z = \frac{6x}{y}
$$

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16) $ A = \frac{L}{2L} $ (solve for $ L $)

Wait — simplify first:
$$
A = \frac{L}{2L} = \frac{1}{2}, \quad \text{for } L \ne 0
$$
So $ A = \frac{1}{2} $. But this implies $ A $ is constant regardless of $ L $, so unless $ A = \frac{1}{2} $, no solution exists.
But if we assume the original was meant to be $ A = \frac{L}{2} $ or similar, perhaps a typo.

Wait — let's double-check:
If $ A = \frac{L}{2L} $, then $ A = \frac{1}{2} $, which means $ A $ is always 0.5, so $ L $ can't be solved uniquely.

But maybe the equation was meant to be $ A = \frac{L}{2} $? That would make sense.

Alternatively, suppose it's:
$$
A = \frac{L}{2L} \Rightarrow A = \frac{1}{2}
$$
Then $ A $ is fixed, so no unique solution for $ L $ unless $ A = \frac{1}{2} $, in which case any $ L \ne 0 $ works.

But likely, this is a typo. Let's suppose it's:
Possibility: $ A = \frac{L}{2} $ → Then $ L = 2A $

Or possibly: $ A = \frac{L}{2} $, but given as $ A = \frac{L}{2L} $, which simplifies to $ A = \frac{1}{2} $

So unless $ A = \frac{1}{2} $, no solution. But if we proceed formally:

$$
A = \frac{L}{2L} = \frac{1}{2}
\Rightarrow \text{No solution unless } A = \frac{1}{2}, \text{ then } L \text{ can be any nonzero number}
$$

But since it's asking to solve for $ L $, and $ A $ is given, we must conclude:

$$
\boxed{A = \frac{1}{2}} \Rightarrow \text{No unique solution for } L
$$

But likely, the intended equation was:
$$
A = \frac{L}{2} \Rightarrow L = 2A
$$

Or perhaps:
$$
A = \frac{L}{2} \Rightarrow L = 2A
$$

Given ambiguity, assume typo and suppose it's $ A = \frac{L}{2} $, then:
$$
L = 2A
$$

But as written: $ A = \frac{L}{2L} $ → $ A = \frac{1}{2} $, so unless $ A = \frac{1}{2} $, no solution.

So best to note: The equation $ A = \frac{L}{2L} $ simplifies to $ A = \frac{1}{2} $, so $ L $ cannot be determined uniquely from $ A $ unless $ A = \frac{1}{2} $.

But if we must solve for $ L $, then:

From $ A = \frac{L}{2L} $, multiply both sides by $ 2L $:
$$
2AL = L
\Rightarrow 2AL - L = 0
\Rightarrow L(2A - 1) = 0
$$

So either $ L = 0 $ or $ 2A - 1 = 0 \Rightarrow A = \frac{1}{2} $

But $ L = 0 $ makes denominator zero in original, so invalid.

Thus, only possible if $ A = \frac{1}{2} $, and then $ L $ can be any nonzero value.

So no unique solution.

But since this seems unlikely, probably a typo. Let’s assume it's:
$$
A = \frac{L}{2} \Rightarrow L = 2A
$$

We’ll go with that for now, noting the issue.

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17) $ A = \frac{a + b + c}{3} $ (solve for $ b $)

Multiply both sides by 3:
$$
3A = a + b + c
$$
Subtract $ a $ and $ c $:
$$
b = 3A - a - c
$$

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18) $ 12x - 4y = 20 $ (solve for $ y $)

Subtract $ 12x $:
$$
-4y = -12x + 20
$$
Divide by $-4$:
$$
y = 3x - 5
$$

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19) $ x = \frac{2y - z}{4} $ (solve for $ z $)

Multiply both sides by 4:
$$
4x = 2y - z
$$
Subtract $ 2y $:
$$
4x - 2y = -z
\Rightarrow z = 2y - 4x
$$

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20) $ P = \frac{R - C}{N} $ (solve for $ R $)

Multiply both sides by $ N $:
$$
PN = R - C
$$
Add $ C $:
$$
R = PN + C
$$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1) | $ T = \frac{P}{IR} $ |
| 2) | $ W = \frac{A}{2} - L $ |
| 3) | $ x = \frac{y + 6}{5} $ |
| 4) | $ y = \frac{2x - 8}{3} $ |
| 5) | $ x = 15 - y $ |
| 6) | $ b = y - mx $ |
| 7) | $ y = \frac{c - ax}{b} $ |
| 8) | $ b = \frac{A}{1.2h} - c $ |
| 9) | $ L = \frac{V}{WH} $ |
| 10) | $ r^2 = \frac{A}{4\pi} $ |
| 11) | $ h = \frac{V}{\pi r^2} $ |
| 12) | $ x = \frac{y + 14}{7} $ |
| 13) | $ y = 2A - x $ |
| 14) | $ I = \frac{E}{R} $ |
| 15) | $ z = \frac{6x}{y} $ |
| 16) | Ambiguous – if $ A = \frac{L}{2L} $, then $ A = \frac{1}{2} $, so $ L $ not uniquely solvable. Likely typo; if $ A = \frac{L}{2} $, then $ L = 2A $ |
| 17) | $ b = 3A - a - c $ |
| 18) | $ y = 3x - 5 $ |
| 19) | $ z = 2y - 4x $ |
| 20) | $ R = PN + C $ |

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