## College Algebra (6th Edition)

Conditional equation. $x=2$
Given equation, $\frac{x+2}{x+3} + \frac{1}{x^{2}+2x-3} - 1 = 0$ By factoring, $x^{2}+2x-3=(x+3)(x-1)$, the given equation can be written as $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} - 1 = 0$ $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1$ $x=1$ or $x=-3$ makes the denominator zero, so, $\frac{x+2}{x+3} + \frac{1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ Taking LCD, $\frac{(x+2)(x-1)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $\frac{(x^{2}+2x-x-2)+1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $\frac{x^{2}+x-1}{(x+3)(x-1)} = 1 ; x\ne 1,-3;$ $x^{2}+x-1 = (x+3)(x-1) ; x\ne 1,-3;$ $x^{2}+x-1 = x^{2}+2x-3 ; x\ne 1,-3;$ $x^{2}+x-1 -x^{2}-2x+3=0 ; x\ne 1,-3;$ $-x+2=0 ; x\ne 1,-3;$ $x=2 ; x\ne 1,-3;$ This equation has one solution, so it is a conditional equation.