I cant seem to find a way to actually find any solutions to this....any suggestions?

$\displaystyle 3/(x^2+x)-1/(x+5)=0$

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- Apr 13th 2010, 09:07 PMEliteCoreSolve
I cant seem to find a way to actually find any solutions to this....any suggestions?

$\displaystyle 3/(x^2+x)-1/(x+5)=0$ - Apr 13th 2010, 09:15 PMharish21
- Apr 13th 2010, 09:28 PMEliteCore
- Apr 13th 2010, 09:31 PMpickslides
Now expand the numerator and group like terms. (Rofl)

- Apr 13th 2010, 09:31 PMharish21
- Apr 13th 2010, 09:36 PMWilmer
If numerator / denominator = 0, then numerator = 0

- Apr 13th 2010, 09:44 PMEliteCore
is $\displaystyle (-(x-5))/(x(x+2))$

the answer? - Apr 13th 2010, 09:48 PMharish21
NO.

$\displaystyle \frac{3(x+5)-x(x+1)}{(x^2+x)(x+5)} $ is what you had gotten.

according to your question, this term = 0, so you have,

$\displaystyle \frac{3(x+5)-x(x+1)}{(x^2+x)(x+5)} = 0$

Look at what**Wilmer**has said in the post above. That gives you:

$\displaystyle 3(x+5)-x(x+1) = 0$

$\displaystyle 3x+15-x^2-x=0$

$\displaystyle 2x+15-x^2=0$

$\displaystyle x^2-2x-15=0$

Solve this quadratic equation for the value of x!