Rational Equations: Solving worksheet with ten problems to solve for variables.

Let's solve each of the rational equations step by step. Rational equations involve fractions with variables in the denominator, so we need to:

1. Identify restrictions (values that make denominators zero).
2. Eliminate denominators by multiplying both sides by the least common denominator (LCD).
3. Solve the resulting equation.
4. Check for extraneous solutions (solutions that make any denominator zero).

---

1) $\frac{3}{x} + \frac{1}{4} = \frac{8}{x}$

Step 1: Restrictions: $x \neq 0$

Step 2: Multiply both sides by $4x$ (LCD of $x$ and $4$):

$$
4x\left(\frac{3}{x} + \frac{1}{4}\right) = 4x\left(\frac{8}{x}\right)
$$

$$
4x \cdot \frac{3}{x} + 4x \cdot \frac{1}{4} = 4x \cdot \frac{8}{x}
$$

$$
12 + x = 32
$$

Step 3: Solve:

$$
x = 32 - 12 = 20
$$

Step 4: Check: $x = 20$ is not restricted → valid.

✔ Answer: $x = 20$

---

2) $\frac{x}{4} = \frac{6}{3}$

Simplify right side:

$$
\frac{x}{4} = 2
$$

Multiply both sides by 4:

$$
x = 8
$$

✔ Answer: $x = 8$

---

3) $\frac{2}{x - 2} = \frac{5}{x - 1}$

Restrictions: $x \neq 2$, $x \neq 1$

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

$$
(x - 2)(x - 1)\left(\frac{2}{x - 2}\right) = (x - 2)(x - 1)\left(\frac{5}{x - 1}\right)
$$

$$
2(x - 1) = 5(x - 2)
$$

$$
2x - 2 = 5x - 10
$$

$$
-2 + 10 = 5x - 2x \Rightarrow 8 = 3x \Rightarrow x = \frac{8}{3}
$$

Check: $x = \frac{8}{3} \neq 1, 2$ → valid.

✔ Answer: $x = \frac{8}{3}$

---

4) $\frac{4}{w - 2} + \frac{3}{w + 2} = \frac{16}{w^2 - 4}$

Note: $w^2 - 4 = (w - 2)(w + 2)$

Restrictions: $w \neq 2$, $w \neq -2$

LCD: $(w - 2)(w + 2)$

Multiply both sides:

$$
(w - 2)(w + 2)\left(\frac{4}{w - 2} + \frac{3}{w + 2}\right) = (w - 2)(w + 2)\left(\frac{16}{(w - 2)(w + 2)}\right)
$$

$$
4(w + 2) + 3(w - 2) = 16
$$

$$
4w + 8 + 3w - 6 = 16 \Rightarrow 7w + 2 = 16
$$

$$
7w = 14 \Rightarrow w = 2
$$

But $w = 2$ makes denominator zero → extraneous solution

✘ No solution

✔ Answer: No solution

---

5) $\frac{x}{5} = \frac{3}{x - 2}$

Restriction: $x \neq 2$

Multiply both sides by $5(x - 2)$:

$$
5(x - 2)\left(\frac{x}{5}\right) = 5(x - 2)\left(\frac{3}{x - 2}\right)
$$

$$
x(x - 2) = 15
$$

$$
x^2 - 2x - 15 = 0
$$

Factor:

$$
(x - 5)(x + 3) = 0 \Rightarrow x = 5 \text{ or } x = -3
$$

Check: $x = 5$: $5 - 2 = 3 \neq 0$ → valid
$x = -3$: $-3 - 2 = -5 \neq 0$ → valid

✔ Answer: $x = 5$ or $x = -3$

---

6) $\frac{x}{2} = \frac{-7}{5 - x}$

Restriction: $x \neq 5$

Multiply both sides by $2(5 - x)$:

$$
2(5 - x)\left(\frac{x}{2}\right) = 2(5 - x)\left(\frac{-7}{5 - x}\right)
$$

$$
x(5 - x) = -14
$$

$$
5x - x^2 = -14
$$

Bring all terms to one side:

$$
-x^2 + 5x + 14 = 0 \Rightarrow x^2 - 5x - 14 = 0
$$

Use quadratic formula:

$$
x = \frac{5 \pm \sqrt{(-5)^2 + 4(1)(14)}}{2} = \frac{5 \pm \sqrt{25 + 56}}{2} = \frac{5 \pm \sqrt{81}}{2} = \frac{5 \pm 9}{2}
$$

$$
x = \frac{14}{2} = 7 \quad \text{or} \quad x = \frac{-4}{2} = -2
$$

Check: $x = 7$: $5 - 7 = -2 \neq 0$ → valid
$x = -2$: $5 - (-2) = 7 \neq 0$ → valid

✔ Answer: $x = 7$ or $x = -2$

---

7) $\frac{2y}{2} = \frac{-4}{y + 4}$

Simplify left side: $\frac{2y}{2} = y$

So:

$$
y = \frac{-4}{y + 4}
$$

Restriction: $y \neq -4$

Multiply both sides by $y + 4$:

$$
y(y + 4) = -4
\Rightarrow y^2 + 4y + 4 = 0
\Rightarrow (y + 2)^2 = 0
\Rightarrow y = -2
$$

Check: $y = -2 \neq -4$ → valid

✔ Answer: $y = -2$

---

8) $\frac{2}{2t - 3} = \frac{3}{3 - 2t}$

Note: $3 - 2t = -(2t - 3)$

So:

$$
\frac{2}{2t - 3} = \frac{3}{-(2t - 3)} = -\frac{3}{2t - 3}
$$

Now:

$$
\frac{2}{2t - 3} = -\frac{3}{2t - 3}
$$

Multiply both sides by $2t - 3$ (≠ 0 → $t \neq \frac{3}{2}$):

$$
2 = -3
$$

✘ Contradiction → No solution

✔ Answer: No solution

---

9) $\frac{5x}{10x + 5} - 4 = \frac{3x - 1}{2x + 1}$

First, simplify left side:

Note: $10x + 5 = 5(2x + 1)$

So:

$$
\frac{5x}{5(2x + 1)} = \frac{x}{2x + 1}
$$

So equation becomes:

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

Restriction: $2x + 1 \neq 0 \Rightarrow x \neq -\frac{1}{2}$

Now multiply both sides by $2x + 1$:

$$
x - 4(2x + 1) = 3x - 1
\Rightarrow x - 8x - 4 = 3x - 1
\Rightarrow -7x - 4 = 3x - 1
$$

$$
-4 + 1 = 3x + 7x \Rightarrow -3 = 10x \Rightarrow x = -\frac{3}{10}
$$

Check: $x = -\frac{3}{10}$, $2x + 1 = 2(-0.3) + 1 = -0.6 + 1 = 0.4 \neq 0$ → valid

✔ Answer: $x = -\frac{3}{10}$

---

10) $\frac{3}{5m} = \frac{m + 3}{7m}$

Restriction: $m \neq 0$

Multiply both sides by $35m$ (LCM of $5m$ and $7m$):

$$
35m \cdot \frac{3}{5m} = 35m \cdot \frac{m + 3}{7m}
\Rightarrow 7 \cdot 3 = 5(m + 3)
\Rightarrow 21 = 5m + 15
\Rightarrow 6 = 5m \Rightarrow m = \frac{6}{5}
$$

Check: $m = \frac{6}{5} \neq 0$ → valid

✔ Answer: $m = \frac{6}{5}$

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✔ Final Answers:

1) $x = 20$
2) $x = 8$
3) $x = \frac{8}{3}$
4) No solution
5) $x = 5$ or $x = -3$
6) $x = 7$ or $x = -2$
7) $y = -2$
8) No solution
9) $x = -\frac{3}{10}$
10) $m = \frac{6}{5}$

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