Question

In the following exercises, solve.

$\frac{q}{3q-9}-\frac{3}{4q+12}=\frac{7q^2+6q+63}{24q^2-216}$

Short answer

The solution is$q=3,-3.$

Step-by-step solution

Step 1. Given Information 

The given rational equation is$\frac{q}{3q-9}-\frac{3}{4q+12}=\frac{7q^2+6q+63}{24q^2-216}$

Step 2. Least Common Denominator(LCD)

On Simplifying the given equation,

$$\begin{gathered} \frac{q}{3q-9}-\frac{3}{4q+12}=\frac{7q^2+6q+63}{24q^2-216} \\ \frac{q}{3(q-3)}-\frac{3}{4(q+3)}=\frac{7q^2+6q+63}{24(q^2-9)} \\ \frac{q}{3(q-3)}-\frac{3}{4(q+3)}=\frac{7q^2+6q+63}{24\left(q+3\right)(q-3)} \end{gathered}$$

The LCD is$24(q+3)(q-3)$

Step 3. Clearing fractions

Multiply all terms by LCD,

$$\begin{gathered} 24(q+3)(q-3)\left[\frac{q}{3(q-3)}-\frac{3}{4(q+3)}\right]=\frac{7q^2+6q+63}{24(q+3)(q-3)}\times 24(q+3)(q-3) \\ 8(q+3)\left(q\right)-6(q-3)\left(3\right)=7q^2+6q+63 \\ 8q^2+24q-18q+54=7q^2+6q+63 \\ {q}^2-9=0 \\ (q+3)(q-3)=0 \end{gathered}$$

Step 4. Solution of the equation

Using the Zero Product Property,

$$\begin{gathered} q+3=0;q-3=0 \\ q=-3,q=3 \end{gathered}$$