# Solve the equation. \frac{5}{2x+3}+\frac{4}{2x-3}=\frac{14x+3}{4x^{2}-9}Z

Solve the equation.
$\frac{5}{2x+3}+\frac{4}{2x-3}=\frac{14x+3}{4{x}^{2}-9}$
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Step 1
We have solve the equation
$\frac{5}{2x+3}+\frac{4}{2x-3}=\frac{14x+3}{4{x}^{2}-9}$
Step 2
Here,
$\frac{5}{2x+3}+\frac{4}{2x-3}=\frac{14x+3}{4{x}^{2}-9}$
$⇒\frac{5\left(2x-3\right)+4\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{14x+3}{4{x}^{2}-9}$
$⇒\frac{10x-15+8x+12}{{\left(2x\right)}^{2}-{\left(3\right)}^{2}}=\frac{14x+3}{4{x}^{2}-9}$
$⇒\frac{18x-3}{4{x}^{2}-9}=\frac{14x+3}{4{x}^{2}-9}$
$⇒18x-3=14x+3$
$⇒18x-14x=3+3$
$⇒4x=6$
$⇒x=\frac{6}{4}=\frac{3}{2}$
$⇒x=\frac{3}{2}$