SOLUTION: Solve: x/(x - 4) - 2/(x + 3) = 20/x^2 - x - 12 >>>( "/" indicates fraction bar, not division). Here is what I have. x(x+3)/(x-4)(x+3) - 2(x-4)/(x-4)(x+3) = 20/(x-4)(x+3)

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve: x/(x - 4) - 2/(x + 3) = 20/x^2 - x - 12 >>>( "/" indicates fraction bar, not division). Here is what I have. x(x+3)/(x-4)(x+3) - 2(x-4)/(x-4)(x+3) = 20/(x-4)(x+3)       Log On


   

Question 796141: Solve: x/(x - 4) - 2/(x + 3) = 20/x^2 - x - 12 >>>( "/" indicates fraction bar, not division).
Here is what I have.
x(x+3)/(x-4)(x+3) - 2(x-4)/(x-4)(x+3) = 20/(x-4)(x+3)
I'm stuck here. Is my numerator of 20 supposed to have the LCD following it also? Ex: 20(x-4)(x+3)/(x-4)(x+3)? Once I get my equation set up to where I can make it equal to zero, I can solve from there.
Thank you for your time!
Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

Solve: x/(x - 4) - 2/(x + 3) = 20/x^2 - x - 12 >>>( "/" indicates fraction bar, not division).
Here is what I have.
x(x+3)/(x-4)(x+3) - 2(x-4)/(x-4)(x+3) = 20/(x-4)(x+3)
I'm stuck here. Is my numerator of 20 supposed to have the LCD following it also? Ex: 20(x-4)(x+3)/(x-4)(x+3)? Once I get my equation set up to where I can make it equal to zero, I can solve from there.
Thank you for your time!

x%2F%28x+-+4%29+-+2%2F%28x+%2B+3%29+=+20%2F%28x%5E2+-+x+-+12%29

x%2F%28x+-+4%29+-+2%2F%28x+%2B+3%29+=+20%2F%28%28x+-+4%29%28x+%2B+3%29%29

x(x + 3) - 2(x - 4) = 20 ----- Multiplying by LCD, (x - 4)(x + 3)

x%5E2+%2B+3x+-+2x+%2B+8+=+20

x%5E2+%2B+x+-+12+=+0

Solve for x, but be reminded that x+%3C%3E+4, and x+%3C%3E+-+3

You can do the check!!

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