Answer:
Step-by-step explanation:
[tex]\frac{3}{x+3}-\frac{4}{x-3}=\frac{3*(x-3)}{(x+3)(x-3)}-\frac{4(x+3)}{(x-3)(x+3)}\\\\=\frac{3*x-3*3}{(x+3)(x-3)}-\frac{4*x+4*3}{(x+3)(x-3)}\\\\=\frac{3x-9}{x^{2}-3^{2}}-\frac{4x+12}{x^{2}-3^{2}}\\\\=\frac{3x - 9 -(4x + 12)}{x^{2}-3^{2}}\\\\=\frac{3x-9-4x-12}{x^{2}-9}\\\\=\frac{-x-21}{x^{2}-9}\\\\[/tex]
[tex]\frac{3}{x+3}-\frac{4}{x-3}=\frac{5x}{x^{2}-9}\\\\\frac{-x-21}{x^{2}-9}=\frac{5x}{x^{2}-9}\\\\-x-21=\frac{5x}{(x^{2}-9)}*(x^{2}-9)\\[/tex]
-x - 21 = 5x
Add 'x' to both sides
-x -21 +x = 5x + x
-21 = 6x
6x = -21
x = -21/6
= -7/2
x = -3.5
