osula9a

2021-12-14

Perform the multiplication or division and simplify.
$\frac{x+3}{4{x}^{2}-9}÷\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$

zesponderyd

Step 1
Simplify: $\frac{x+3}{4{x}^{2}-9}÷\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$
Step 2
Inverting and changing to multiplication:
$\frac{x+3}{4{x}^{2}-9}÷\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$
$=\frac{x+3}{4{x}^{2}-9}×\frac{2{x}^{2}+7x-15}{{x}^{2}+7x+12}$
$=\frac{x+3}{{\left(2x\right)}^{2}-{3}^{2}}×\frac{2{x}^{2}+10x-3x-15}{{x}^{2}+4x+3x+12}$
$=\frac{x+3}{\left(2x-3\right)\left(2x+3\right)}×\frac{2x\left(x+5\right)-3\left(x+5\right)}{x\left(x+4\right)+3\left(x+4\right)}$
$=\frac{x+3}{\left(2x-3\right)\left(2x+3\right)}×\frac{\left(2x-3\right)\left(x+5\right)}{\left(x+3\right)\left(x+4\right)}$
Step 3
cancelling the same terms, we get:
$=\frac{\left(x+3\right)}{\left(2x-3\right)\left(2x+3\right)}×\frac{\left(2x-3\right)\left(x+5\right)}{\left(x+3\right)\left(x+4\right)}$
$=\frac{x+5}{\left(2x+3\right)\left(x+4\right)}$

Lindsey Gamble

Step 1
Given: $\frac{x+3}{4{x}^{2}-9}÷\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$
Divide $\frac{x+3}{4{x}^{2}-9}$ by $\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$ by multiplying $\frac{x+3}{4{x}^{2}-9}$ by the reciprocal of $\frac{{x}^{2}+7x+12}{2{x}^{2}+7x-15}$
$\frac{\left(x+3\right)\left(2{x}^{2}+7x-15\right)}{\left(4{x}^{2}-9\right)\left({x}^{2}+7x+12\right)}$
Factor the expressions that are not already factored.
$\frac{\left(2x-3\right)\left(x+3\right)\left(x+5\right)}{\left(2x-3\right)\left(x+3\right)\left(x+4\right)\left(2x+3\right)}$
Cancel out $\left(2x-3\right)\left(x+3\right)$ in both numerator and denominator.
$\frac{x+5}{\left(x+4\right)\left(2x+3\right)}$
Expand the expression.
$\frac{x+5}{2{x}^{2}+11x+12}$

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