Question

(x/x+3)-(8/x+2)=(-13x-6)/(x^2+5x+6)​

Answer 1

Answer:

x = -9

x = 2

Step-by-step explanation:

Given:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{x^2+5x+6}$

First, note that

$x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)$

Since $x+2$ and $x+3$ stay in the denominator, then

$x\neq -2\ \text{and}\ x\neq -3$

Now add two fractions which stay in the left part. The common denominator is $(x+2)(x+3),$ so multiply the numerator of the first fraction by $(x+2)$ and the numerator of the second fraction by $(x+3)$ and subtract them in the numerator:

$\dfrac{x}{x+3}-\dfrac{8}{x+2}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\\dfrac{x(x+2)-8(x+3)}{(x+2)(x+3)}=\dfrac{-13x-6}{(x+2)(x+3)}\\ \\x(x+2)-8(x+3)=-13x-6\\ \\x^2+2x-8x-24=-13x-6\\ \\x^2+2x-8x+13x-24+6=0\\ \\x^2+7x-18=0\\ \\x^2+9x-2x-18=0\\ \\x(x+9)-2(x+9)=0\\ \\(x+9)(x-2)=0\\ \\x+9=0\ \text{or}\ x-2=0\\ \\x=-9\ \text{or}\ x=2$