1Cademy - Solving $$1 - \frac{5}{y} = -\frac{6}{y^2}$$

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Strategy to Solve Equations with Rational Expressions

Example

Solving $1 - \frac{5}{y} = -\frac{6}{y^2}$

Solve the rational equation $1 - \frac{5}{y} = -\frac{6}{y^2}$ by applying the five-step strategy for equations with rational expressions. This example demonstrates a case where clearing fractions produces a quadratic equation rather than a linear one.

Step 1 — Identify restricted values. The denominators $y$ and $y^2$ are both zero when $y = 0$ , so record the restriction $y \neq 0$ .

Step 2 — Find the LCD. The denominators in the equation are $1$ (implicit), $y$ , and $y^2$ . The LCD is $y^2$ .

Step 3 — Clear the fractions. Multiply both sides by $y^2$ and distribute:

$y^2 \cdot 1 - y^2 \cdot \frac{5}{y} = y^2 \cdot \left(-\frac{6}{y^2}\right)$

Simplify each term: $y^2 \cdot 1 = y^2$ , $y^2 \cdot \frac{5}{y} = 5y$ , and $y^2 \cdot \frac{6}{y^2} = 6$ . The fraction-free equation is:

$y^2 - 5y = -6$

Step 4 — Solve the resulting equation. Write in standard form by adding $6$ to both sides:

$y^2 - 5y + 6 = 0$

Factor the trinomial: $(y - 2)(y - 3) = 0$ . Apply the Zero Product Property:

$y - 2 = 0 \quad \text{or} \quad y - 3 = 0$

$y = 2 \quad \text{or} \quad y = 3$

Step 5 — Check. Neither $2$ nor $3$ equals the restricted value $0$ , so neither is extraneous.

For $y = 2$ : $1 - \frac{5}{2} = -\frac{3}{2}$ and $-\frac{6}{4} = -\frac{3}{2}$ , so $-\frac{3}{2} = -\frac{3}{2}$ ✓

For $y = 3$ : $1 - \frac{5}{3} = -\frac{2}{3}$ and $-\frac{6}{9} = -\frac{2}{3}$ , so $-\frac{2}{3} = -\frac{2}{3}$ ✓

The solutions are $y = 2$ and $y = 3$ . Unlike the earlier example $\frac{1}{x} + \frac{1}{3} = \frac{5}{6}$ , which produced a linear equation after clearing fractions, this equation has the variable in the denominator raised to a power — clearing fractions with the LCD of $y^2$ produces the quadratic $y^2 - 5y + 6 = 0$ , which requires factoring and the Zero Product Property to find two solutions.

Updated 2026-04-21

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OpenStax Elementary Algebra

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