Question

In the following exercises, solve.

$\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{x^2-16}$

Short answer

The solution of the equation $\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{x^2-16}$is

$x=\frac{-4}{5}$

Step-by-step solution

Step 1. Given information  

The given equation is

$\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{x^2-16}$

Step 2, Variable values for which denominators are zero

Factor all the denominators of the equation

$$\begin{gathered} \frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{x^2-16} \\ \frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{\left(x+4\right)\left(x-4\right)} \end{gathered}$$

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

so$\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{\left(x+4\right)\left(x-4\right)},x=4,x=-4$

Step 3. Solution of equation 

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

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

role="math" localid="1647802390820" $$\begin{gathered} (x+4)(x-4)\left(\frac{3}{x+4}+\frac{7}{x-4}\right)=(x+4)(x-4)\left(\frac{8}{x^2-16}\right) \\ (x+4)(x-4)\left(\frac{3}{x+4}\right)+(x+4)(x-4)\left(\frac{7}{x-4}\right)=(x+4)(x-4)\left(\frac{8}{x^2-16}\right) \\ (x-4)\left(3\right)+(x+4)\left(7\right)=8 \\ 3x-12+7x+28=8 \\ 10x=-8 \\ x=\frac{-4}{5} \end{gathered}$$

so the solution of the equation isrole="math" localid="1647802536523" $x=\frac{-4}{5}$

Step 4. Verification 

Substitute $x=\frac{-4}{5}$in the equation

$$\begin{gathered} \frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{x^2-16} \\ \frac{3}{\left(\frac{-4}{5}\right)+4}+\frac{7}{\left(\frac{-4}{5}\right)-4}\stackrel{?}{=}\frac{8}{{\left(\frac{-4}{5}\right)}^2-16} \\ \frac{3·5}{-4+20}+\frac{7·5}{-4-20}\stackrel{?}{=}\frac{8·25}{16-16·25} \\ \frac{15}{16}-\frac{35}{24}\stackrel{?}{=}-\frac{25}{48} \\ -\frac{25}{48}=-\frac{25}{48} \end{gathered}$$

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