Fractional equation reducing to a quadratic

Text

Solve the following fractional equation: $\frac{x-3}{x + 3}-1 = \frac{x-1}{x + 1}+2.$

Solution

Existence conditions: $x\neq -3$ and $x\neq -1$. Simplify the two sides: $\frac{x-3}{x+3}-1 = \frac{(x-3)-(x+3)}{x+3} = \frac{-6}{x+3}, \qquad \frac{x-1}{x+1}+2 = \frac{(x-1)+2(x+1)}{x+1} = \frac{3x+1}{x+1}.$ The equation becomes $\dfrac{-6}{x+3} = \dfrac{3x+1}{x+1}$; cross-multiplying: $-6(x+1) = (3x+1)(x+3) \;\Rightarrow\; -6x-6 = 3x^2+10x+3 \;\Rightarrow\; 3x^2+16x+9 = 0.$ $x = \frac{-16 \pm \sqrt{256-108}}{6} = \frac{-16 \pm 2\sqrt{37}}{6} = \frac{-8 \pm \sqrt{37}}{3} \approx -0.639 \ \text{and}\ -4.694.$ Both acceptable (different from $-3$ and $-1$). $\boxed{x = \frac{-8 \pm \sqrt{37}}{3}}$