## Algebra and Trigonometry 10th Edition

x =$\frac{7}{4}$
Solve : $\frac{1}{x-2} + \frac{3}{x+3} = \frac{4}{x^{2} + x -6}$ $\frac{1}{x-2} + \frac{3}{x+3} = \frac{4}{(x-2)(x+3)}$ Multiplying both sides with (x-2)(x+3) x+3 + 3(x-2) = 4 x+3+3x-6 = 4 4x = 7 x =$\frac{7}{4}$