Free Printable Solving Literal Equations Worksheets - Free Printable

It looks like you've uploaded a Literal Equations Worksheet with the answers already provided, but you're asking to solve the problem and explain the solution.

Since the worksheet shows both the problems and their solutions, I'll go through each problem step by step, explaining how to solve literal equations (equations with multiple variables) for a specific variable. This will help clarify the reasoning behind each answer.

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🔹 What is a Literal Equation?

A literal equation is an equation that involves several letters (variables). We solve it by isolating one variable in terms of the others.

Let’s walk through each problem:

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1) $ 2(x + 4y) = 18 $, solve for $ x $

Step-by-step:

1. Divide both sides by 2:
$$
x + 4y = 9
$$

2. Subtract $ 4y $ from both sides:
$$
x = 9 - 4y
$$

✔ Answer: $ x = 9 - 4y $

---

2) $ \frac{2xy}{3} = 5z $, solve for $ x $

Step-by-step:

1. Multiply both sides by 3 to eliminate the denominator:
$$
2xy = 15z
$$

2. Divide both sides by $ 2y $:
$$
x = \frac{15z}{2y}
$$

✔ Answer: $ x = \frac{15z}{2y} $

---

3) $ R = \frac{l + 3w}{2} $, solve for $ w $

Step-by-step:

1. Multiply both sides by 2:
$$
2R = l + 3w
$$

2. Subtract $ l $ from both sides:
$$
2R - l = 3w
$$

3. Divide by 3:
$$
w = \frac{2R - l}{3}
$$

✔ Answer: $ w = \frac{2R - l}{3} $

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4) $ h = vt - 16t^2 $, solve for $ v $

Step-by-step:

1. Add $ 16t^2 $ to both sides:
$$
h + 16t^2 = vt
$$

2. Divide both sides by $ t $:
$$
v = \frac{h + 16t^2}{t}
$$

✔ Answer: $ v = \frac{h + 16t^2}{t} $

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5) $ C = \frac{5}{9}(F - 32) $, solve for $ F $

This is the Celsius to Fahrenheit conversion formula.

Step-by-step:

1. Multiply both sides by $ \frac{9}{5} $:
$$
\frac{9}{5}C = F - 32
$$

2. Add 32 to both sides:
$$
F = \frac{9}{5}C + 32
$$

✔ Answer: $ F = \frac{9}{5}C + 32 $

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6) $ A = p(1 + rt) $, solve for $ t $

Step-by-step:

1. Divide both sides by $ p $:
$$
\frac{A}{p} = 1 + rt
$$

2. Subtract 1 from both sides:
$$
\frac{A}{p} - 1 = rt
$$

3. Divide by $ r $:
$$
t = \frac{\frac{A}{p} - 1}{r} = \frac{A - p}{pr}
$$

✔ Answer: $ t = \frac{A - p}{pr} $

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7) $ A = \frac{1}{2h(b + c)} $, solve for $ b $

Wait — this seems to have a typo or formatting issue. The original equation is likely:
$$
A = \frac{1}{2h(b + c)}
$$
But that would make $ A $ a fraction involving $ h, b, c $. However, the answer suggests it's meant to be:
$$
A = \frac{1}{2h(b + c)} \quad \text{or perhaps} \quad A = \frac{1}{2h(b + c)} \Rightarrow \text{but this doesn't match typical area formulas.}
$$

Actually, looking at the answer:
$$
b = \frac{1 - 2Ahc}{2Ah}
$$
That suggests the original equation might be:
$$
A = \frac{1}{2h(b + c)}
\Rightarrow \text{Then } \frac{1}{A} = 2h(b + c)
\Rightarrow \frac{1}{2Ah} = b + c
\Rightarrow b = \frac{1}{2Ah} - c
$$
But that doesn’t match the given answer.

Wait — let’s double-check.

The correct form might actually be:
$$
A = \frac{1}{2h(b + c)} \quad \text{(unlikely)}
$$
But more plausibly, the intended equation is:
$$
A = \frac{1}{2h(b + c)} \quad \text{is not standard.}
$$

Alternatively, maybe it's:
$$
A = \frac{1}{2h(b + c)} \Rightarrow \text{no.}
$$

Wait — the answer says:
$$
b = \frac{1 - 2Ahc}{2Ah}
$$

Let’s reverse-engineer:

Suppose:
$$
A = \frac{1}{2h(b + c)}
\Rightarrow \frac{1}{A} = 2h(b + c)
\Rightarrow \frac{1}{2Ah} = b + c
\Rightarrow b = \frac{1}{2Ah} - c
$$

But that’s not matching.

Wait — perhaps the original equation is:
$$
A = \frac{1}{2h(b + c)} \quad \text{is incorrect.}
$$

Looking again: The image shows:
$$
A = \frac{1}{2h(b + c)}
$$

But that would mean:
$$
A = \frac{1}{2h(b + c)}
\Rightarrow \frac{1}{A} = 2h(b + c)
\Rightarrow \frac{1}{2Ah} = b + c
\Rightarrow b = \frac{1}{2Ah} - c
$$

Still not matching.

But the answer is:
$$
b = \frac{1 - 2Ahc}{2Ah}
$$

So let’s suppose the original equation was:
$$
A = \frac{1}{2h(b + c)} \quad \text{is wrong.}
$$

Wait — maybe it's:
$$
A = \frac{1}{2h(b + c)} \quad \text{no.}
$$

Wait — could it be:
$$
A = \frac{1}{2h(b + c)} \quad \text{no.}
$$

Alternatively, perhaps the equation is:
$$
A = \frac{1}{2h(b + c)} \quad \text{is not correct.}
$$

Let’s try to derive from the answer.

Given:
$$
b = \frac{1 - 2Ahc}{2Ah}
$$

Multiply numerator and denominator:
$$
b = \frac{1}{2Ah} - \frac{2Ahc}{2Ah} = \frac{1}{2Ah} - c
\Rightarrow b + c = \frac{1}{2Ah}
\Rightarrow 2Ah(b + c) = 1
\Rightarrow A = \frac{1}{2h(b + c)}
$$

Yes! So the original equation is:
$$
A = \frac{1}{2h(b + c)}
$$

So solving:
1. Take reciprocal of both sides:
$$
\frac{1}{A} = 2h(b + c)
$$

2. Divide by $ 2h $:
$$
\frac{1}{2Ah} = b + c
$$

3. Subtract $ c $:
$$
b = \frac{1}{2Ah} - c = \frac{1 - 2Ahc}{2Ah}
$$

✔ Answer: $ b = \frac{1 - 2Ahc}{2Ah} $

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8) $ A = \frac{x + y + z}{2} $, solve for $ y $

Step-by-step:

1. Multiply both sides by 2:
$$
2A = x + y + z
$$

2. Subtract $ x $ and $ z $:
$$
y = 2A - x - z
$$

✔ Answer: $ y = 2A - x - z $

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✔ Summary of All Solutions:

| Problem | Equation | Solved for | Answer |
|--------|---------|------------|--------|
| 1 | $ 2(x + 4y) = 18 $ | $ x $ | $ x = 9 - 4y $ |
| 2 | $ \frac{2xy}{3} = 5z $ | $ x $ | $ x = \frac{15z}{2y} $ |
| 3 | $ R = \frac{l + 3w}{2} $ | $ w $ | $ w = \frac{2R - l}{3} $ |
| 4 | $ h = vt - 16t^2 $ | $ v $ | $ v = \frac{h + 16t^2}{t} $ |
| 5 | $ C = \frac{5}{9}(F - 32) $ | $ F $ | $ F = \frac{9}{5}C + 32 $ |
| 6 | $ A = p(1 + rt) $ | $ t $ | $ t = \frac{A - p}{pr} $ |
| 7 | $ A = \frac{1}{2h(b + c)} $ | $ b $ | $ b = \frac{1 - 2Ahc}{2Ah} $ |
| 8 | $ A = \frac{x + y + z}{2} $ | $ y $ | $ y = 2A - x - z $ |

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📌 Key Tips for Solving Literal Equations:

- Isolate the target variable using inverse operations.
- Treat other variables as constants.
- Use multiplication/division to eliminate denominators or coefficients.
- Distribute if needed.
- Combine like terms carefully.

If you have any specific question about a problem or need a visual explanation, feel free to ask!