SOLUTION: Solve the equation by factoring. (2x/x+3)+(5/x)-4=(18/x^2+3x) ----------- ----------- -------- I have gotten this far with the problem.... LCM = X(x+3) thus I have 2x^2+5x+15

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve the equation by factoring. (2x/x+3)+(5/x)-4=(18/x^2+3x) ----------- ----------- -------- I have gotten this far with the problem.... LCM = X(x+3) thus I have 2x^2+5x+15      Log On


   

Question 570645: Solve the equation by factoring.
(2x/x+3)+(5/x)-4=(18/x^2+3x)
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I have gotten this far with the problem....
LCM = X(x+3) thus I have 2x^2+5x+15-4x^2-12=18--->-2x^2+5x+3=18--->-2x^2+5x-15=0

I think my problem is the (-2x^2) which I got from 2x^2-4x^2 and the (-4x^2) is from multiplying the LCM (x(x+3))to -4.
Could you show me where my error is?
Thank you
Kenneth
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
%282x%29%2F%28x%2B3%29%2B5%2Fx+-4+=+18%2F%28x%5E2%2B3x%29 Simplify. The LCD isx%5E2%2B3x
so we have:
-2x%5E2-7x%2B15+=+18 Subtract 18 from both sides.
-2x%5E2-7x-3+=+0 Multiply through by -1 to get:
2x%5E2%2B7x%2B3+=+0 Factor.
%282x%2B1%29%28x%2B3%29+=+0 Apply the zero product rule:
2x%2B1+=+0 or x%2B3+=+0 so...
x+=+-1%2F2 or x+=+-3
Notice: The left side becomes...
%282x%28x%29%2B5%28x%2B3%29-4%28x%28x%2B3%29%29%29%2Fx%28x%2B3%29 which, when simplified becomes:
%282x%5E2%2B5x%2B15-4x%5E2-12x%29%2Fx%28x%2B3%29=%28-2x%5E2-7x%2B15%29%2F%28x%5E2%2B3x%29