# Solve

• Apr 13th 2010, 09:07 PM
EliteCore
Solve
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 PM
harish21
Quote:

Originally Posted by EliteCore
I cant seem to find a way to actually solve this....any suggestions?

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

Where are you stuck at? its easy.

$\displaystyle \frac{a}{b} - \frac{c}{d} = \frac{(a\times d) - (b \times c)}{b \times d}$

• Apr 13th 2010, 09:28 PM
EliteCore
Quote:

Originally Posted by harish21
Where are you stuck at? its easy.

$\displaystyle \frac{a}{b} - \frac{c}{d} = \frac{(a\times d) - (b \times c)}{b \times d}$

Then I get this:

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

can I go any further?
• Apr 13th 2010, 09:31 PM
pickslides
Now expand the numerator and group like terms. (Rofl)
• Apr 13th 2010, 09:31 PM
harish21
Quote:

Originally Posted by EliteCore
Then I get this:

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

can I go any further?

now expand the numerator by multiplying the terms
• Apr 13th 2010, 09:36 PM
Wilmer
If numerator / denominator = 0, then numerator = 0
• Apr 13th 2010, 09:44 PM
EliteCore
is $\displaystyle (-(x-5))/(x(x+2))$

• Apr 13th 2010, 09:48 PM
harish21
Quote:

Originally Posted by EliteCore
is $\displaystyle (-(x-5))/(x(x+2))$

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!