Answer
$$x= 4$$
Work Step by Step
We have $$\frac{2x}{x^2+6x+8} = \frac{x}{x+4} - \frac{2}{x+2}$$
As we see we have a rational expression so we need to multiply each side of the equation by the minimum common multiple of the denominators.
Note that $(x+4)(x+2) = x^2 + 6x + 8$ so the minimun common multiple of the denominators is $(x+4)(x+2)$. Multiplying each side of the equation by mcm $$(x+4)(x+2)(\frac{2x}{x^2+6x+8}) = (x+4)(x+2)(\frac{x}{x+4} - \frac{2}{x+2})$$ $$\frac{(x+4)(x+2)(2x)}{x^2+6x+8} = \frac{(x+4)(x+2)(x)}{x+4} - \frac{(x+4)(x+2)(2)}{x+2}$$
Simplifying each rational expression $$2x = (x+2)(x) - (x+4)(2)$$
Doing all the multiplications in order to eliminate the parenthesis $$2x = (x)(x) + (2)(x) - (x)(2) - (4)(2)$$ $$2x = x^2 + 2x -2x - 8$$
Taking all the terms to the left side of the equation and equalizing to $0$
$$2x - x^2 - 2x + 2x + 8 =0$$
Simplifying $$-x^2 +2x +8 = 0$$
Multiplying each side by $-1$ $$x^2 -2x -8 = 0$$
Note we can factorize the expression $$(x-4)(x+2) = 0$$
Equalizing each parenthesis to $0$ we can get our two values.
For the first one $x-4 = 0$ so $x_{1}= 4$
For the second one $x+2 = 0$ so $x_{2} = -2$
Substituting each one of the value $x = 4$ results to be a solution but $x=-2$ not. So the only solution for this equation is $x = 4$.
