Math worksheet for solving rational equations, including problems with fractions and variables in denominators, designed for practice and skill development.

Let's solve each of the rational equations step by step, and check for extraneous solutions (solutions that make a denominator zero, which are not valid).

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1. $\frac{1}{x} = \frac{6}{5x} + 1$

Step 1: Eliminate denominators by multiplying both sides by the LCD, which is $5x$:

$$
5x \cdot \left( \frac{1}{x} \right) = 5x \cdot \left( \frac{6}{5x} + 1 \right)
$$

$$
5 = 6 + 5x
$$

Step 2: Solve for $x$:

$$
5 - 6 = 5x \Rightarrow -1 = 5x \Rightarrow x = -\frac{1}{5}
$$

Check for extraneous solutions:

- $x = -\frac{1}{5}$ → Not zero → Valid.

✔ Solution: $x = -\frac{1}{5}$

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2. $1 = \frac{2}{r^2} - \frac{1}{r}$

Step 1: Multiply both sides by the LCD, which is $r^2$:

$$
r^2 \cdot 1 = r^2 \cdot \left( \frac{2}{r^2} - \frac{1}{r} \right)
$$

$$
r^2 = 2 - r
$$

Step 2: Rearrange into standard form:

$$
r^2 + r - 2 = 0
$$

Factor:

$$
(r + 2)(r - 1) = 0
\Rightarrow r = -2 \text{ or } r = 1
$$

Check for extraneous solutions:

- $r = -2$: Denominators $r^2 = 4$, $r = -2$ → No division by zero → Valid.
- $r = 1$: $r^2 = 1$, $r = 1$ → Valid.

✔ Solutions: $r = -2$, $r = 1$

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3. $\frac{2}{3}x - \frac{5}{6} = \frac{3}{4}$

This is not a rational equation in the same sense — it's linear with fractions. But we'll solve it anyway.

Step 1: Eliminate fractions by multiplying all terms by the LCD of 3, 6, 4 → 12:

$$
12 \cdot \left( \frac{2}{3}x - \frac{5}{6} \right) = 12 \cdot \frac{3}{4}
$$

$$
8x - 10 = 9
$$

Step 2: Solve:

$$
8x = 19 \Rightarrow x = \frac{19}{8}
$$

No denominators with variables → no extraneous solutions.

✔ Solution: $x = \frac{19}{8}$

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4. $\frac{x}{x - 1} - \frac{1}{x - 2} = \frac{11}{x^2 - 3x + 2}$

Step 1: Factor the denominator on the right:

$$
x^2 - 3x + 2 = (x - 1)(x - 2)
$$

So LCD is $(x - 1)(x - 2)$

Multiply both sides by $(x - 1)(x - 2)$:

$$
(x - 1)(x - 2) \left( \frac{x}{x - 1} - \frac{1}{x - 2} \right) = (x - 1)(x - 2) \cdot \frac{11}{(x - 1)(x - 2)}
$$

Simplify:

$$
x(x - 2) - 1(x - 1) = 11
$$

$$
x^2 - 2x - x + 1 = 11
\Rightarrow x^2 - 3x + 1 = 11
\Rightarrow x^2 - 3x - 10 = 0
$$

Factor:

$$
(x - 5)(x + 2) = 0 \Rightarrow x = 5 \text{ or } x = -2
$$

Check for extraneous solutions:

- $x = 5$: denominators $x - 1 = 4$, $x - 2 = 3$ → OK
- $x = -2$: $x - 1 = -3$, $x - 2 = -4$ → OK

✔ Solutions: $x = 5$, $x = -2$

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5. $\frac{p - 4}{5p} = \frac{1}{5p} + 1$

Step 1: Multiply both sides by $5p$ (LCD):

$$
5p \cdot \left( \frac{p - 4}{5p} \right) = 5p \cdot \left( \frac{1}{5p} + 1 \right)
$$

$$
p - 4 = 1 + 5p
$$

Step 2: Solve:

$$
p - 4 = 1 + 5p \Rightarrow -4 - 1 = 5p - p \Rightarrow -5 = 4p \Rightarrow p = -\frac{5}{4}
$$

Check for extraneous solutions:

- $p = -\frac{5}{4} \neq 0$ → No division by zero → Valid.

✔ Solution: $p = -\frac{5}{4}$

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6. $\frac{x}{x + 4} = 3 - \frac{4}{x + 4}$

Step 1: LCD is $x + 4$. Multiply both sides:

$$
(x + 4) \cdot \left( \frac{x}{x + 4} \right) = (x + 4) \cdot \left( 3 - \frac{4}{x + 4} \right)
$$

$$
x = 3(x + 4) - 4
\Rightarrow x = 3x + 12 - 4
\Rightarrow x = 3x + 8
\Rightarrow -2x = 8 \Rightarrow x = -4
$$

Check for extraneous solutions:

- $x = -4$ → denominator $x + 4 = 0$ → undefined

✘ Extraneous solution!

So, no solution.

🚫 Answer: No solution

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7. $x + \frac{6}{x - 3} = \frac{2x}{x - 3}$

Step 1: LCD is $x - 3$. Multiply both sides:

$$
(x - 3) \cdot \left( x + \frac{6}{x - 3} \right) = (x - 3) \cdot \left( \frac{2x}{x - 3} \right)
$$

$$
x(x - 3) + 6 = 2x
\Rightarrow x^2 - 3x + 6 = 2x
\Rightarrow x^2 - 5x + 6 = 0
$$

Factor:

$$
(x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3
$$

Check for extraneous solutions:

- $x = 2$: $x - 3 = -1$ → OK
- $x = 3$: $x - 3 = 0$ → division by zero → ✘ Extraneous

So only $x = 2$ is valid.

✔ Solution: $x = 2$

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8. $\frac{1}{x^2} + \frac{4}{x} = \frac{3}{x^2}$

Step 1: LCD is $x^2$. Multiply both sides:

$$
x^2 \cdot \left( \frac{1}{x^2} + \frac{4}{x} \right) = x^2 \cdot \frac{3}{x^2}
$$

$$
1 + 4x = 3
\Rightarrow 4x = 2 \Rightarrow x = \frac{1}{2}
$$

Check for extraneous solutions:

- $x = \frac{1}{2} \neq 0$ → No division by zero → Valid.

✔ Solution: $x = \frac{1}{2}$

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## ✔ Final Answers:

1. $x = -\frac{1}{5}$
2. $r = -2, 1$
3. $x = \frac{19}{8}$
4. $x = 5, -2$
5. $p = -\frac{5}{4}$
6. No solution
7. $x = 2$
8. $x = \frac{1}{2}$

> 🔍 Note: Always check for values that make any denominator zero — those are extraneous and must be rejected.