Answer
$\left\{-\dfrac{11}{7},0\right\}$
Work Step by Step
Multiplying by the $LCD=
2(x+1)(x+2)
,$ the given equation $
\dfrac{2}{x+1}+\dfrac{3}{x+2}=\dfrac{7}{2}
,$ is equivalent to
\begin{align*}\require{cancel}
2(x+1)(x+2)\left(\dfrac{2}{x+1}+\dfrac{3}{x+2}\right)&=\left(\dfrac{7}{2}\right)2(x+1)(x+2)
\\\\
2(x+2)(2)+2(x+1)(3)&=7(x+1)(x+2)
\\
4x+8+6x+6&=7(x^2+3x+2)
&(\text{use }(a+b)(c+d)=ac+ad+bc+bd)
\\
10x+14&=7x^2+21x+14
\\
0&=7x^2+(21x-10x)+(14-14)
\\
0&=7x^2+11x
\\
7x^2+11x&=0
.\end{align*}
Factoring the $GCF=x,$ the equation above is equivalent to
\begin{align*}
x(7x+11)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x=0 & 7x+11=0
\\
& 7x=-11
\\\\
& x=-\dfrac{11}{7}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=0: & \text{If }x=-\dfrac{11}{7}:
\\\\
\dfrac{2}{0+1}+\dfrac{3}{0+2}\overset{?}=\dfrac{7}{2} &
\dfrac{2}{-\frac{11}{7}+1}+\dfrac{3}{-\frac{11}{7}+2}\overset{?}=\dfrac{7}{2}
\\\\
\dfrac{2}{1}+\dfrac{3}{2}\overset{?}=\dfrac{7}{2} &
\dfrac{2}{-\frac{11}{7}+\frac{7}{7}}+\dfrac{3}{-\frac{11}{7}+\frac{14}{7}}\overset{?}=\dfrac{7}{2}
\\\\
\dfrac{4}{2}+\dfrac{3}{2}\overset{?}=\dfrac{7}{2} &
\dfrac{2}{-\frac{4}{7}}+\dfrac{3}{\frac{3}{7}}\overset{?}=\dfrac{7}{2}
\\\\
\dfrac{7}{2}\overset{\checkmark}=\dfrac{7}{2} &
2\div\left(-\dfrac{4}{7}\right)+3\div\dfrac{3}{7}\overset{?}=\dfrac{7}{2}
\\\\
&
2\cdot\left(-\dfrac{7}{4}\right)+3\cdot\dfrac{7}{3}\overset{?}=\dfrac{7}{2}
\\\\
&
\cancelto12\cdot\left(-\dfrac{7}{\cancelto24}\right)+\cancel3\cdot\dfrac{7}{\cancel3}\overset{?}=\dfrac{7}{2}
\\\\
&
-\dfrac{7}{2}+7\overset{?}=\dfrac{7}{2}
\\\\
&
-\dfrac{7}{2}+\dfrac{14}{2}\overset{?}=\dfrac{7}{2}
\\\\
&
\dfrac{7}{2}\overset{\checkmark}=\dfrac{7}{2}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
\dfrac{2}{x+1}+\dfrac{3}{x+2}=\dfrac{7}{2}
$ is $\left\{-\dfrac{11}{7},0\right\}$.
