3 years ago

3x+6/x^2-x-6+2x/x^2+x-12

1 answer:

6 0

$\dfrac{3x+6}{x^2-x-6} + \dfrac{2x}{x^2+x-12}$

$= \dfrac{3(x+2)}{(x + 2)(x - 3)} + \dfrac{2x}{ (x - 3)(x + 4)}$

$= \dfrac{3}{(x - 3)} + \dfrac{2x}{ (x -3)(x + 4)}$

$= \dfrac{3(x+4)}{(x - 3)(x + 4)} + \dfrac{2x}{ (x -3 )(x + 4)}$

$= \dfrac{3(x+4) + 2x }{(x - 3)(x + 4)}$

$= \dfrac{3x+12 + 2x }{(x - 3)(x + 4)}$

$= \dfrac{5x+12}{(x - 3)(x + 4)}$

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Simplify the following expression completely, where x is greater than or equal too 0.

Dennis_Churaev [7]

First you want the numbers under the radical to be the same so you can combine like terms.

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$\sqrt{405x^3y^4} = \sqrt{5x*81x^2y^4} =9xy^2 \sqrt{5x}$
$\sqrt{80x^3y^4} = \sqrt{5x*16x^2y^4} = 4xy^2 \sqrt{5x}$

Now add together to get $14xy^2 \sqrt{5x}$