Solve Rational Equations and Inequalities (solutions, examples ... - Free Printable

Great! Let's go through the example shown in your image and solve it step-by-step, following the five steps for solving rational equations.

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🔷 Problem:

Solve the rational equation:

$$
\frac{1}{x} + \frac{1}{x - 8} = \frac{1}{3}
$$

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✔ Step 1: Factor all the denominators

The denominators are:
- $ x $
- $ x - 8 $
- $ 3 $

These are already factored. So we proceed.

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✔ Step 2: Find the least common denominator (LCD)

We need the LCD of:
- $ x $
- $ x - 8 $
- $ 3 $

Since these have no common factors, the LCD is:
$$
\text{LCD} = 3x(x - 8)
$$

✔ This matches what’s shown in the example.

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✔ Step 3: Multiply each side by the LCD to cancel out denominators

Multiply both sides of the equation by $ 3x(x - 8) $:

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

Now distribute the LCD to each term:

$$
3x(x - 8) \cdot \frac{1}{x} + 3x(x - 8) \cdot \frac{1}{x - 8} = 3x(x - 8) \cdot \frac{1}{3}
$$

Simplify each term:

- $ 3x(x - 8)/x = 3(x - 8) $
- $ 3x(x - 8)/(x - 8) = 3x $
- $ 3x(x - 8)/3 = x(x - 8) $

So the equation becomes:

$$
3(x - 8) + 3x = x(x - 8)
$$

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✔ Step 4: Solve the equation

Expand both sides:

Left side:
$$
3x - 24 + 3x = 6x - 24
$$

Right side:
$$
x^2 - 8x
$$

Now write the equation:

$$
6x - 24 = x^2 - 8x
$$

Bring all terms to one side:

$$
0 = x^2 - 8x - 6x + 24
\Rightarrow x^2 - 14x + 24 = 0
$$

Factor the quadratic:

Look for two numbers that multiply to 24 and add to -14 → $-12$ and $-2$

$$
(x - 12)(x - 2) = 0
$$

So the solutions are:

$$
x = 12 \quad \text{or} \quad x = 2
$$

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✔ Step 5: Check the solutions

We must ensure that no solution makes any denominator zero.

Original equation has denominators:
- $ x $
- $ x - 8 $

So check:
- If $ x = 12 $: $ x = 12 \neq 0 $, $ x - 8 = 4 \neq 0 $ → ✔ Valid
- If $ x = 2 $: $ x = 2 \neq 0 $, $ x - 8 = -6 \neq 0 $ → ✔ Valid

Now plug into original equation to verify:

#### Check $ x = 12 $:

$$
\frac{1}{12} + \frac{1}{12 - 8} = \frac{1}{12} + \frac{1}{4} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3} \quad ✔
$$

#### Check $ x = 2 $:

$$
\frac{1}{2} + \frac{1}{2 - 8} = \frac{1}{2} + \frac{1}{-6} = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \quad ✔
$$

Both solutions work!

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✔ Final Answer:

$$
\boxed{x = 12 \quad \text{or} \quad x = 2}
$$

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📌 Summary of Steps:

1. Factor denominators.
2. Find LCD → $ 3x(x - 8) $
3. Multiply both sides by LCD to eliminate denominators.
4. Solve resulting equation → quadratic → factor.
5. Check solutions → both valid.

✔ Both $ x = 12 $ and $ x = 2 $ are valid solutions.

Let me know if you'd like a different example or explanation!