Answer:
-2 is an extraneous solution because it creates 0 in the denominator.
There is one acceptable solution to the equation
Step-by-step explanation:
x² + 5x + 6 = x² + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
[tex]\frac{x}{x+2}+\frac{2}{x^{2}+5x+6}=\frac{5}{(x+3)}\\\\\frac{x}{x+2}+\frac{2}{(x+2)(x+3)}=\frac{5}{(x+3)}\\\\\frac{x*(x+3)}{(x+2)(x+3)}+\frac{2}{(x+2)(x+3)}=\frac{5*(x+2)}{(x+3)(x+2)}\\\\\frac{x^{2}+3x+2}{(x+2)(x+3)}=\frac{5x+10}{(x+3)(x+2)}\\\\[/tex]
x² + 3x + 2 = 5x + 10
x² + 3x - 5x + 2 - 10 = 0
x² - 2x - 8 = 0
x² + 2x - 4x - 8 = 0
x(x + 2) - 4(x + 2) = 0
(x + 2)(x - 4) = 0
x +2 = 0 or x - 4 = 0
x = - 2 or x = 4
