Example

Solving 1x+13=56\frac{1}{x} + \frac{1}{3} = \frac{5}{6}

Solve the rational equation 1x+13=56\frac{1}{x} + \frac{1}{3} = \frac{5}{6} by applying the five-step strategy for equations with rational expressions.

Step 1 — Identify restricted values. The fraction 1x\frac{1}{x} has the variable xx in its denominator. If x=0x = 0, this expression is undefined, so record the restriction x0x \neq 0.

Step 2 — Find the LCD. The denominators in the equation are xx, 33, and 66. The least common denominator of these three is 6x6x.

Step 3 — Clear the fractions. Multiply both sides of the equation by 6x6x and use the Distributive Property:

6x1x+6x13=6x566x \cdot \frac{1}{x} + 6x \cdot \frac{1}{3} = 6x \cdot \frac{5}{6}

Simplify each term: 6x1x=66x \cdot \frac{1}{x} = 6, 6x13=2x6x \cdot \frac{1}{3} = 2x, and 6x56=5x6x \cdot \frac{5}{6} = 5x. The equation is now fraction-free:

6+2x=5x6 + 2x = 5x

Step 4 — Solve the resulting equation. Subtract 2x2x from both sides: 6=3x6 = 3x. Divide both sides by 33: x=2x = 2.

Step 5 — Check. The value x=2x = 2 does not equal the restricted value 00, so it is not extraneous. Substitute x=2x = 2 into the original equation:

12+13=36+26=56\frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

Since 56=56\frac{5}{6} = \frac{5}{6} is true, x=2x = 2 is confirmed as the solution.

This example demonstrates the complete five-step strategy applied to a rational equation whose variable appears in a denominator. Unlike clearing fractions in a linear equation — where all denominators are constants — here the LCD contains the variable xx, making the restriction in Step 1 essential. The process of multiplying by the LCD of 6x6x eliminates all fractions in one step, converting the rational equation into the simple linear equation 6+2x=5x6 + 2x = 5x.

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Updated 2026-04-21

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