In the following exercises, solve.

$\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{y^2-81}$

The given equation is

$\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{y^2-81}$

Factor all the denominators of the equation

$$\begin{gathered} \frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{y^2-81} \\ \frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{\left(y-9\right)\left(y+9\right)} \end{gathered}$$

Terms of the equation are not defined for $x=9\&x=-9$

sorole="math" localid="1647805523118" $\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{\left(y-9\right)\left(y+9\right)},x\ne 9,x\ne -9$

The least common denominator of all denominators in the equation is$\left(y-9\right)\left(y+9\right)$

multiply both sides of the equation by the least common denominator$\left(y-9\right)\left(y+9\right)$

role="math" localid="1647807550424" $$\begin{gathered} \left(y-9\right)\left(y+9\right)\left(\frac{5}{y-9}+\frac{1}{y+9}\right)=\left(y-9\right)\left(y+9\right)\left(\frac{18}{y^2-81}\right) \\ \left(y+9\right)\left(5\right)+\left(y-9\right)\left(1\right)=18 \\ 5y+45+y-9=18 \\ 6y=-18 \\ y=-3 \end{gathered}$$

so the solution of the equation isrole="math" localid="1647807562610" $y=-3$

Substitute $y=-3$in the equation

$$\begin{gathered} \frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{y^2-81} \\ \frac{5}{-3-9}+\frac{1}{-3+9}\stackrel{?}{=}\frac{18}{{\left(-3\right)}^2-81} \\ \frac{5}{-12}+\frac{1}{6}\stackrel{?}{=}\frac{18}{9-81} \\ -\frac{5}{12}+\frac{2}{12}\stackrel{?}{=}-\frac{18}{72} \\ -\frac{3}{12}\stackrel{?}{=}-\frac{1}{4} \\ -\frac{1}{4}=-\frac{1}{4} \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution is correct.