SOLUTION: hello, i need help solving this equation and simplying my answer as much as possible, thanks so much! 2x 3 1 -------

SOLUTION: hello, i need help solving this equation and simplying my answer as much as possible, thanks so much! 2x 3 1 ------- - ------ + ------ x^2+x-6 x+

Question 57032: hello, i need help solving this equation and simplying my answer as much as possible, thanks so much!
2x 3 1
------- - ------ + ------
x^2+x-6 x+3 x-2
Found 2 solutions by Cintchr, BGluvsmath:
Answer by Cintchr(481) (Show Source): You can put this solution on YOUR website!

Factor the denominator of the first fraction

Multiply the 2nd fraction by and the 3rd by

With the denominators all the same ... combine the numerators

Distibute through the numerator

Combine the X terms in the numerator

Combine the constants

Answer by BGluvsmath(64) (Show Source): You can put this solution on YOUR website!
2x/(x^2+x-6) - 3/(x+3) + 1/(x-2)
To get a common denominator (necessary for adding and subtracting fractions), multiply the 2nd term by x-2 and the 3rd term by x+3:
2x/(x^2+x-6) - 3(x-2)/((x+3)(x-2)) + 1(x+3)/((x-2)(x+3))
(x+3)(x-2) simplifies to x^2+x-6, so:
2x-3(x-2)+(x+3)/(x^2+x-6)
2x-3x+6+x+3/x^2+x-6
The x's cancel out of the numerator, so this simplifies to:
9/(x^2+x-6)
RELATED QUESTIONS
Hello, I need help with this problem and simplyfying my answer as much as possible. Thank (answered by stanbon)
i need help solving this equation {{{ log(2,x)+log(2,x-6)=4 }}} and finding the domain... (answered by solver91311)
I need help solving these two equations: 5/6(x-1)-2/3(2x+1)=3 x/2 + 1/x-1 = x/x-1 (answered by richwmiller)
Hi! I need some help solving this equation and if possible expressing my answer as on... (answered by tutorcecilia)
Hi! I need some help with this problem and simplifying my answer as much as possible,... (answered by Cintchr)
Hello, I need help with the following problems as I get an error message on my... (answered by ankor@dixie-net.com)
Hi! Could you please help me solve this equation and simplify my answer as much as... (answered by stanbon)
HeLLo! And if you are helping me.... I truly appreciate it! Im having a hard time with... (answered by stanbon)
Solve for the variable in each equation below. Show all your work. Provide a check for... (answered by sudhanshu_kmr)