1. ## Equation

Can someone help me with this one?

$\displaystyle \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0$

2. $\displaystyle \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0$

$\displaystyle \frac {5(x-4)-1(x+2)}{(x+2)(x-4)} - \frac{5}{2x+4} = 0$

$\displaystyle \frac {5x-20-x-2}{(x+2)(x-4)}-\frac{5}{2x+4} = 0$

$\displaystyle \frac {4x-22}{(x+2)(x-4)}-\frac{5}{2x+4} = 0$

$\displaystyle \frac {(4x-22)(2x+4)-5(x+2)(x-4)}{(x+2)(x-4)(2x+4)}= 0$ (cross multiply)

(4x-22)(2x+4)-5(x+2)(x-4) = 0
8x^2+16x-44x-88-5(x^2-4x+2x-8) = 0
8x^2+16x-44x-88-5x^2+10x+40 = 0
3x^2-18x-48 = 0
(x-8)(3x+6) = 0
x=8 or x = -2

Hope it helps.

Originally Posted by hkerbest
Can someone help me with this one?

$\displaystyle \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0$

3. Originally Posted by hkerbest
Can someone help me with this one?

$\displaystyle \frac {5}{x+2} - \frac{1}{x-4} -\frac{5}{2x+4} = 0$
first get the lcd that is $\displaystyle 2(x+2)(x-2)$

$\displaystyle \frac{5*2(x-2)-1(2)-5(x-2)}{2(x+2)(x-2)}$

$\displaystyle \frac{10x-20- 2 -5x+10}{2(x+2)(x-2)}$

$\displaystyle \frac{5x-8}{2(x-4)}$

4. Originally Posted by princess_21
first get the lcd that is $\displaystyle 2(x+2)(x-2)$

$\displaystyle \frac{5*2(x-2)-1(2)-5(x-2)}{2(x+2)(x-2)}$

$\displaystyle \frac{10x-20- 2 -5x+10}{2(x+2)(x-2)}$

$\displaystyle \frac{5x-8}{2(x-4)}$

Sorry i don't think this is right. As the answer you got is x = 8/5, when you resubstitute it back to the solution. You will not get back zero. I will format my previous solution using latex shortly.

5. right, oops my mistake..