Answer
$\left\{-8,\dfrac{9}{11}\right\}$
Work Step by Step
Multiplying by the $LCD=
4m(m+9)
,$ the given equation $
\dfrac{2}{m}+\dfrac{3}{m+9}=\dfrac{11}{4}
,$ is equivalent to
\begin{align*}\require{cancel}
4m(m+9)\left(\dfrac{2}{m}+\dfrac{3}{m+9}\right)&=\left(\dfrac{11}{4}\right)4m(m+9)
\\\\
4(m+9)(2)+4m(3)&=11(m)(m+9)
\\
8m+72+12m&=11m^2+99m
\\
0&=11m^2+(99m-8m-12m)-72
\\
0&=11m^2+79m-72
\\
11m^2+79m-72&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(11m-9)(m+8)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
11m-9=0 & m+8=0
\\
11m=9 & m=-8
\\\\
m=\dfrac{9}{11}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }m=\dfrac{9}{11}: & \text{If }m=-8:
\\\\
\dfrac{2}{\frac{9}{11}}+\dfrac{3}{\frac{9}{11}+9}\overset{?}=\dfrac{11}{4} &
\dfrac{2}{-8}+\dfrac{3}{-8+9}\overset{?}=\dfrac{11}{4}
\\\\
2\div\dfrac{9}{11}+\dfrac{3}{\frac{9}{11}+\frac{99}{11}}\overset{?}=\dfrac{11}{4} &
-\dfrac{1}{4}+\dfrac{3}{1}\overset{?}=\dfrac{11}{4}
\\\\
2\cdot\dfrac{11}{9}+\dfrac{3}{\frac{108}{11}}\overset{?}=\dfrac{11}{4} &
-\dfrac{1}{4}+\dfrac{12}{4}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{22}{9}+3\div\dfrac{108}{11}\overset{?}=\dfrac{11}{4} &
\dfrac{11}{4}\overset{\checkmark}=\dfrac{11}{4}
\\\\
\dfrac{22}{9}+3\cdot\dfrac{11}{108}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{22}{9}+\cancelto13\cdot\dfrac{11}{\cancelto{36}{108}}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{22}{9}+\dfrac{11}{36}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{22}{9}\cdot\dfrac{4}{4}+\dfrac{11}{36}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{88}{36}+\dfrac{11}{36}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{99}{36}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{\cancelto{11}{99}}{\cancelto{4}{36}}\overset{?}=\dfrac{11}{4}
\\\\
\dfrac{11}{4}\overset{\checkmark}=\dfrac{11}{4}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
\dfrac{2}{m}+\dfrac{3}{m+9}=\dfrac{11}{4}
$ is $\left\{-8,\dfrac{9}{11}\right\}$.
