Solving Rational Equations

Strategy: clear the denominators

Multiply every term by the LCD to eliminate all fractions, then solve the resulting polynomial equation. Always check for extraneous solutions — values that make any original denominator zero.

Step-by-step: Solve $\dfrac{1}{x} + \dfrac{1}{x+2} = \dfrac{5}{8}$

Step 1 — Find the LCD. Denominators: $x$, $(x+2)$, $8$. LCD $= 8x(x+2)$.

Step 2 — Multiply every term by the LCD:

$8x(x+2) \cdot \frac{1}{x} + 8x(x+2) \cdot \frac{1}{x+2} = 8x(x+2) \cdot \frac{5}{8}$

$8(x+2) + 8x = 5x(x+2)$

Step 3 — Expand and simplify:

$8x + 16 + 8x = 5x^2 + 10x$ $16x + 16 = 5x^2 + 10x$ $5x^2 - 6x - 16 = 0$

Step 4 — Solve using the quadratic formula: $a = 5, b = -6, c = -16$.

$x = \frac{6 \pm \sqrt{36 + 320}}{10} = \frac{6 \pm \sqrt{356}}{10}$

$x_1 \approx 2.487, \quad x_2 \approx -1.287$

Step 5 — Check for extraneous solutions. Domain restrictions: $x \neq 0$ and $x \neq -2$.

$x_1 = 2.487 \neq 0$ and $\neq -2$ ✓. $x_2 = -1.287 \neq 0$ and $\neq -2$ ✓. Both are valid.

Why extraneous solutions appear

Multiplying by the LCD can introduce solutions that make the LCD zero (i.e., make a denominator zero). These must be rejected.

Try it in the visualization

The graph shows $y = 1/x + 1/(x+2)$ and $y = 5/8$. Intersections are the solutions. Vertical asymptotes at $x = 0$ and $x = -2$ show the excluded values.