Answer
$x=\dfrac{13}{3}$
Work Step by Step
$\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3x+3}$
First, we can take out common factor $3$ in the denominator of the fraction $\dfrac{1}{3x+3}$
$\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3(x+1)}$
We multiply the whole equation by $6(x+1)$, which is the least common multiple of the denominators present in the equation. We do this to avoid working with fractions.
$6(x+1)[\dfrac{3}{x+1}-\dfrac{1}{2}=\dfrac{1}{3(x+1)}]$
Evaluate the products and simplify:
$\dfrac{18(x+1)}{x+1}-\dfrac{6(x+1)}{2}=\dfrac{6(x+1)}{3(x+1)}$
$18-3(x+1)=2$
$18-3x-3=2$
Solve for $x$:
$-3x=2-18+3$
$-3x=-13$
$x=\dfrac{-13}{-3}=\dfrac{13}{3}$
