[tex]\frac{3}{x-2} + \frac{8}{x + 3} = 2[/tex] ; restriction is x ≠ 2, x ≠ -3
⇒ [tex](\frac{x+3}{x+3} )\frac{3}{x - 2} + (\frac{x - 2}{x - 2})\frac{8}{x + 3} = (\frac{(x + 3)(x - 2)}{(x + 3)(x - 2)} ) 2[/tex]
⇒ 3(x + 3) + 8(x - 2) = 2(x + 3)(x - 2)
⇒ 3x + 9 + 8x - 16 = 2(x² + x - 6)
⇒ 3x + 9 + 8x - 16 = 2x² + 2x - 12
⇒ 11x - 7 = 2x² + 2x - 12
⇒ 0 = 2x² - 9x - 5
⇒ 0 = (2x + 1)(x - 5)
0 = 2x + 1 , 0 = x - 5
-1 = 2x 5 = x
[tex]-\frac{1}{2} = x[/tex]
Answer: {[tex]-\frac{1}{2}[/tex], 5}
