## Algebra and Trigonometry 10th Edition

Solve : $\frac{6 }{x} - \frac{2}{x+3} = \frac{3(x+5)}{x^{2}+3x}$ Multiply by $(x^{2}+3x)$ on both sides as $x^{2}+3x$ = x(x+3) 6(x+3) - 2(x) = 3(x+5) 6x + 18 - 2x = 3x + 15 4x - 3x = 15 - 18 x = -3 Plug x=-3 into the expression and the denominator becomes zero and since division by zero is undefined, so x=-3 is extraneous.