Pls help in a middle of an exam

Mathematics

1 answer:

KIM [24]2 years ago

3 0

Answer:

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

Step-by-step explanation:

$\frac{2x}{4x^2-9} + \frac{3}{2x^2+x-3}$ can be simplified to:
$\frac{2x}{(2x-3)(2x+3)}+\frac{3}{(2x+3)(x-1)}$

Make them have common denominators so:
$\frac{(2x)(x-1) +3(2x-3)}{(2x-3)(2x+3)(x-1)}$

Simplify:
$\frac{2x^2-2x+6x-9}{(2x-3)(2x+3)(x-1)} = \frac{2x^2+4x-9}{(2x-3)(2x+3)(x-1)}$

I don't think you can simplify that fraction further so that's the answer.

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