In the following exercises, solve.

$\frac{1}{q+4}-\frac{2}{q-2}=1$

The given equation is

$\frac{1}{q+4}-\frac{2}{q-2}=1$

Terms of the equation are not defined for $q=-4\&q=2$

so role="math" localid="1647811126978" $\frac{1}{q+4}-\frac{2}{q-2}=1,q\ne -4,q\ne 2$

The least common denominator of all denominators in the equation is$(q+4)(q-2)$

multiply both sides of the equation by the least common denominator $(q+4)(q-2)$

role="math" localid="1647810979299" $$\begin{gathered} (q+4)(q-2)\left(\frac{1}{q+4}-\frac{2}{q-2}\right)=(q+4)(q-2)\left(1\right) \\ (q-2)\left(1\right)-(q+4)2=(q+4)(q-2) \\ q-2-2q-8=q^2-2q+4q-8 \\ {q}^2+3q+2=0 \end{gathered}$$

Factorize the equation

$$\begin{gathered} q^2+3q+2=0 \\ {q}^2+2q+q+2=0 \\ q(q+2)+1(q+2)=0 \\ (q+1)(q+2)=0 \end{gathered}$$

Use the Zero Product Property

$$\begin{gathered} q+1=0 \\ q=-1 \end{gathered}$$

or $$\begin{gathered} q+2=0 \\ q=-2 \end{gathered}$$

So solutions of the equation are$q=-1\&-2$

Substitute $q=-1$in the equation

role="math" localid="1647811266113" $$\begin{gathered} \frac{1}{q+4}-\frac{2}{q-2}=1 \\ \frac{1}{-1+4}-\frac{2}{-1-2}\stackrel{?}{=}1 \\ \frac{1}{3}+\frac{2}{3}\stackrel{?}{=}1 \\ 1=1 \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution $q=-1$is correct.

Substitute $q=-2$in the equation

$$\begin{gathered} \frac{1}{q+4}-\frac{2}{q-2}=1 \\ \frac{1}{-2+4}-\frac{2}{-2-2}\stackrel{?}{=}1 \\ \frac{1}{2}+\frac{2}{4}\stackrel{?}{=}1 \\ 1=1 \end{gathered}$$

The left-hand side is equal to the right-hand side so the solution $q=-2$is correct.