Marvin Mccormick

2021-02-01

$\frac{3}{2x+3}-1\left(2x-3\right)=\frac{4}{4{x}^{2}-9}$

estenutC

We are given: $\frac{3}{2x+3}-1\left(2x-3\right)=\frac{4}{4{x}^{2}-9}$
Factor $4{x}^{2}-9$: $\frac{3}{2x+3}-1\left(2x-3\right)=\frac{4}{\left(2x+3\right)\left(2x-3\right)}$
We note that $x=±3/2$ are excluded values based on the denominators.
Multiply both sides by the LCD which is $\left(2x+3\right)\left(2x-3\right)3\left(2x-3\right)-\left(2x+3\right)=4$
$6x-9-2x-3=4$
$4x-12=4$
Add 12 to both sides: $4x=16$
Divide both sides by 4: $x=4$

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