Question

-4/(9-x^2) - [(2x+1)/(x^2-3x)]

Answer 1

g to www.wolframalpha.com

Answer 2

i know the answer just not how to get

Answer 3

umm u can get steps in that:)

Answer 4

\[-\frac{4}{9 - x^{2}} - \frac{2x + 1}{x^{2} - 3x} = -\frac{4}{(3 + x)(3 - x)} - \frac{2x + 1}{x(x - 3)} = \]\[\frac{4}{(x + 3)(x - 3)} - \frac{2x + 1}{x(x - 3)} = \frac{4x}{x(x + 3)(x - 3)} - \frac{(2x + 1)(x + 3)}{x(x + 3)(x - 3)} = \]\[\text{Drop the denominators.}\]\[2x = 2x^{2} + 7x + 3\]\[2x^{2} + 5x + 3 = 0\]\[2x^{2} + 2x + 3x + 3 = 0\]\[(2x^{2} + 2x) + (3x + 3) = 0\]\[2x(x + 1) + 3(x + 1) = 0\]\[(2x + 3)(x + 1) = 0\]\[x = -\frac{3}{2} \space and \space -1\]

Answer 5

lol wow

Answer 6

@Calcmathlete deserves a medal:)

Answer 7

lol When people ask for the process, I do not mess around ;)

Answer 8

haha thank you very much calcmathlete! :)