Answer
$\left\{0,\dfrac{59}{13}\right\}$
Work Step by Step
Multiplying by the $LCD=
15(3-p)(5-p)
,$ the given equation $
\dfrac{4}{3-p}+\dfrac{2}{5-p}=\dfrac{26}{15}
,$ is equivalent to
\begin{align*}\require{cancel}
15(3-p)(5-p)\left(\dfrac{4}{3-p}+\dfrac{2}{5-p}\right)&=\left(\dfrac{26}{15}\right)15(3-p)(5-p)
\\\\
15(5-p)(4)+15(3-p)(2)&=26(3-p)(5-p)
\\
300-60p+90-30p&=26(15-8p+p^2)
\\
390-90p&=390-208p+26p^2
\\
0&=26p^2+(-208p+90p)+(390-390)
\\
0&=26p^2-118p
\\\\
\dfrac{0}{2}&=\dfrac{26p^2-118p}{2}
\\\\
0&=13p^2-59p
\\
13p^2-59p&=0
.\end{align*}
Factoring the $GCF=p,$ the equation above is equivalent to
\begin{align*}
p(13p-59)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
p=0 & 13p-59=0
\\
& 13p=59
\\\\
& p=\dfrac{59}{13}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }p=0: & \text{If }p=\dfrac{59}{13}:
\\\\
\dfrac{4}{3-0}+\dfrac{2}{5-0}\overset{?}=\dfrac{26}{15} &
\dfrac{4}{3-\frac{59}{13}}+\dfrac{2}{5-\frac{59}{13}}\overset{?}=\dfrac{26}{15}
\\\\
\dfrac{4}{3}+\dfrac{2}{5}\overset{?}=\dfrac{26}{15} &
\dfrac{4}{\frac{39}{13}-\frac{59}{13}}+\dfrac{2}{\frac{65}{13}-\frac{59}{13}}\overset{?}=\dfrac{26}{15}
\\\\
\dfrac{4}{3}\cdot\dfrac{5}{5}+\dfrac{2}{5}\cdot\dfrac{3}{3}\overset{?}=\dfrac{26}{15} &
\dfrac{4}{-\frac{20}{13}}+\dfrac{2}{\frac{6}{13}}\overset{?}=\dfrac{26}{15}
\\\\
\dfrac{20}{15}+\dfrac{6}{15}\overset{?}=\dfrac{26}{15} &
4\cdot\left(-\dfrac{13}{20}\right)+2\cdot\dfrac{13}{6}\overset{?}=\dfrac{26}{15}
\\\\
\dfrac{26}{15}\overset{\checkmark}=\dfrac{26}{15} &
-\dfrac{13}{5}+\dfrac{13}{3}\overset{?}=\dfrac{26}{15}
\\\\&
-\dfrac{13}{5}\cdot\dfrac{3}{3}+\dfrac{13}{3}\cdot\dfrac{5}{5}\overset{?}=\dfrac{26}{15}
\\\\&
-\dfrac{39}{15}+\dfrac{65}{15}\overset{?}=\dfrac{26}{15}
\\\\&
\dfrac{26}{15}\overset{\checkmark}=\dfrac{26}{15}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
\dfrac{4}{3-p}+\dfrac{2}{5-p}=\dfrac{26}{15}
$ is $\left\{0,\dfrac{59}{13}\right\}$.
