Solving linear equations

• Nov 6th 2008, 12:06 PM
natali.ferno
Solving linear equations
Hello,

hoping some maths genius on here can help me out!

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

i know that i have to factor the bottom $\displaystyle \frac{2}{x-3} + \frac{1}{x+2} = \frac{7}{(x+2) (x-3)}$ were do i go from here?
• Nov 6th 2008, 12:17 PM
clic-clac
Hi,
first you can say that $\displaystyle -2,3$ can't be chosen as solutions.
Then multiply your equation by $\displaystyle (x+2)(x-3)$, so you get a second order equation you can solve.

Edit: first order equation not second...
• Nov 6th 2008, 12:32 PM
natali.ferno
Quote:

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

Okay so I multiplied each term by $\displaystyle (x+2)(x+3)$ which gave me $\displaystyle 2x + 4 + x -3 =7$ than I just solve for x (right?)
• Nov 6th 2008, 12:40 PM
natali.ferno
is my equation right? cause i keep getting the wrong answer when i solve it!!
• Nov 6th 2008, 12:50 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 $\displaystyle x=2$
• Nov 6th 2008, 12:55 PM
clic-clac
You found $\displaystyle 2x+4+x-3=7$, that's correct and becomes $\displaystyle x=2$.
So $\displaystyle 2$ is your solution (because it's different from $\displaystyle -2$ and $\displaystyle 3$).