Answer
$x=\dfrac{33}{6}$
Work Step by Step
$\dfrac{4}{x-1}+\dfrac{2}{x+1}=\dfrac{35}{x^{2}-1}$
The denominator $x^{2}-1$ is a difference of two squares. We can factor it like this.
$\dfrac{4}{x-1}+\dfrac{2}{x+1}=\dfrac{35}{(x-1)(x+1)}$
Now, we can take $(x-1)(x+1)$ to multiply to the left side of the equation:
$(x-1)(x+1)(\dfrac{4}{x-1}+\dfrac{2}{x+1})=35$
Evaluate the products:
$\dfrac{4(x-1)(x+1)}{x-1}+\dfrac{2(x-1)(x+1)}{x+1}=35$
Simplify:
$4(x+1)+2(x-1)=35$
$4x+4+2x-2=35$
Solve for $x$:
$6x=35-4+2$
$6x=33$
$x=\dfrac{33}{6}$
