Answer:
x = -9
x = 2
Step-by-step explanation:
Given:
[tex]\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{x^2+5x+6}[/tex]
First, note that
[tex]x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)[/tex]
Since [tex]x+2[/tex] and [tex]x+3[/tex] stay in the denominator, then
[tex]x\neq -2\ \text{and}\ x\neq -3[/tex]
Now add two fractions which stay in the left part. The common denominator is [tex](x+2)(x+3),[/tex] so multiply the numerator of the first fraction by [tex](x+2)[/tex] and the numerator of the second fraction by [tex](x+3)[/tex] and subtract them in the numerator:
[tex]\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\\dfrac{x(x+2)-8(x+3)}{(x+2)(x+3)}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\x(x+2)-8(x+3)=-13x-6\\ \\x^2+2x-8x-24=-13x-6\\ \\x^2+2x-8x+13x-24+6=0\\ \\x^2+7x-18=0\\ \\x^2+9x-2x-18=0\\ \\x(x+9)-2(x+9)=0\\ \\(x+9)(x-2)=0\\ \\x+9=0\ \text{or}\ x-2=0\\ \\x=-9\ \text{or}\ x=2[/tex]
