Solving linear equations

• Nov 6th 2008, 01:06 PM
natali.ferno
Solving linear equations
Hello,

hoping some maths genius on here can help me out!

$\frac{2}{x-3} + \frac{1}{x+2} = \frac{7}{x^2-x-6}$

i know that i have to factor the bottom $\frac{2}{x-3} + \frac{1}{x+2} = \frac{7}{(x+2) (x-3)}$ were do i go from here?
• Nov 6th 2008, 01:17 PM
clic-clac
Hi,
first you can say that $-2,3$ can't be chosen as solutions.
Then multiply your equation by $(x+2)(x-3)$, so you get a second order equation you can solve.

Edit: first order equation not second...
• Nov 6th 2008, 01:32 PM
natali.ferno
Quote:

Originally Posted by clic-clac
Hi,
first you can say that $-2,3$ can't be chosen as solutions.
Then multiply your equation by $(x+2)(x-3)$, so you get a second order equation you can solve.

Okay so I multiplied each term by $(x+2)(x+3)$ which gave me $2x + 4 + x -3 =7$ than I just solve for x (right?)
• Nov 6th 2008, 01:40 PM
natali.ferno
is my equation right? cause i keep getting the wrong answer when i solve it!!
• Nov 6th 2008, 01:50 PM
masters
Quote:

Originally Posted by natali.ferno
is my equation right? cause i keep getting the wrong answer when i solve it!!

Your equation looks fine. What answer did you get? I got $x=2$
• Nov 6th 2008, 01:55 PM
clic-clac
You found $2x+4+x-3=7$, that's correct and becomes $x=2$.
So $2$ is your solution (because it's different from $-2$ and $3$).