Q32. - Algebra 1

Question

Find each product or quotient.

$\frac{2a^2+7a-15}{a+5}÷\frac{9a^2-4}{3a+2}$

Step-by-Step Solution

Answer

The simplified expression is $\frac{2a-3}{3a-2}$.

Step 1. Define the concept.

Multiplying rational expression: - Let a, b, c, and d be polynomials with $b\ne 0$ and $d\ne 0$. Then $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$.

 

Dividing rational expression: - Let a, b, c, and d be polynomials with $b\ne 0$, $c\ne 0$ and $d\ne 0$. Then $\frac{a}{b}÷\frac{c}{d}=\frac{a}{b}\cdot \frac{d}{c}=\frac{ad}{bc}$.

Step 2. Multiply by the reciprocal of 9 a 2 − 4 3 a + 2 .

First multiply by the reciprocal of $\frac{9a^2-4}{3a+2}$ and then factor the expressions in numerator and denominator.

$$\begin{gathered} \frac{2a^2+7a-15}{a+5}÷\frac{9a^2-4}{3a+2}=\frac{2a^2+7a-15}{a+5}\times \frac{3a+2}{9a^2-4} \\ =\frac{\left(2a-3\right)\left(a+5\right)}{a+5}\times \frac{3a+2}{\left(3a+2\right)\left(3a-2\right)} \end{gathered}$$

Step 3. Simplify the expression.

Further simplify the expression by cancelling out the common factors $a+5$ and $3a+2$.

$$\begin{gathered} \frac{2a^2+7a-15}{a+5}÷\frac{9a^2-4}{3a+2}=\frac{\left(2a-3\right)\left(\cancel{a+5}\right)}{\cancel{a+5}}\times \frac{\cancel{3a+2}}{\left(\cancel{3a+2}\right)\left(3a-2\right)} \\ =\frac{2a-3}{3a-2} \end{gathered}$$

Therefore, the simplified expression is $\frac{2a-3}{3a-2}$.