In the following exercises, solve.

$\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{z^2-100}$

The given equation is

$\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{z^2-100}$

Factor all the denominators of the equation

$$\begin{gathered} \frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{z^2-100} \\ \frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{\left(z-10\right)\left(z+10\right)} \end{gathered}$$

Terms of the equation are not defined for $z=10\&z=-10$

so$\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{z^2-100},z=10,z=-10$

The least common denominator of all denominators in the equation is$\left(z-10\right)\left(z+10\right)$

multiply both sides of the equation by the least common denominatorrole="math" localid="1647808188851" $\left(z-10\right)\left(z+10\right)$

role="math" localid="1647808369722" $$\begin{gathered} \left(z-10\right)\left(z+10\right)\left(\frac{8}{z-10}+\frac{7}{z+10}\right)=\left(z-10\right)\left(z+10\right)\left(\frac{5}{z^2-100}\right) \\ \left(z+10\right)\left(8\right)+\left(z-10\right)\left(7\right)=5 \\ 8z+80+7z-70=5 \\ 15z=-5 \\ z=\frac{-1}{3} \end{gathered}$$

so the solution of the equation is$z=-\frac{1}{3}$

Substitute $z=\frac{-1}{3}$in the equation

$$\begin{gathered} \frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{z^2-100} \\ \frac{8}{\left(\frac{-1}{3}\right)-10}+\frac{7}{\left(\frac{-1}{3}\right)+10}\stackrel{?}{=}\frac{5}{{\left(\frac{-1}{3}\right)}^2-100} \\ \frac{8·3}{-1-30}+\frac{7·3}{-1+30}\stackrel{?}{=}\frac{5·9}{1-900} \\ -\frac{24}{31}+\frac{21}{29}\stackrel{?}{=}-\frac{45}{899} \\ -\frac{45}{899}=-\frac{45}{899} \end{gathered}$$

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