SOLUTION: Hello! I've tried solving this problem multiple times but I'm not sure if I've gotten the right answer or if I was doing it right. Solve: {{{(7x+1)/(x^2-8x+15)+3x/(x-5)=-1/(x-3

Algebra ->  Rational-functions -> SOLUTION: Hello! I've tried solving this problem multiple times but I'm not sure if I've gotten the right answer or if I was doing it right. Solve: {{{(7x+1)/(x^2-8x+15)+3x/(x-5)=-1/(x-3      Log On


   

Question 1035556: Hello! I've tried solving this problem multiple times but I'm not sure if I've gotten the right answer or if I was doing it right.
Solve: %287x%2B1%29%2F%28x%5E2-8x%2B15%29%2B3x%2F%28x-5%29=-1%2F%28x-3%29
This is my work: First off, I marked that x cannot equal 5 or 3, because of how(x-3) and (x-5) are in the denominator and I can't have 0 in the denominator. I turned it into %287x%2B1%29%2F%28x-5%29%28x-3%29%2B3x%2F%28x-5%29=-1%28x-3%29 I noticed that after I factored out x%5E2-8x%2B15 I could rationalize the denominator to (x-5)(x-3) by multiplying 3x/(x-5) with (x-3) and -1/(x-3) with (x-5) and cancel it all out to just end up with %287x%2B1%29%2B3x%28x-3%29=-1%28x-5%29. Next, I simplified it down to 7x%2B1%2B3x%5E2-9x=-x%2B5 to 3x%5E2-x-4=0. I didn't know if I could factor it out anymore mentally, so I chose my safest bet which was using the quadratic formula to get x=4/3 and x=-1. Those two answers would work because it's not 5 or 3 (because then it would equal 0). Am I right? Or did I do something wrong? I hope I was clear enough and I would really appreciate the help!
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
Answer by ikleyn(52775) About Me  (Show Source):
You can put this solution on YOUR website!
.
Hello! I've tried solving this problem multiple times but I'm not sure if I've gotten the right answer or if I was doing it right.
Solve: %287x%2B1%29%2F%28x%5E2-8x%2B15%29%2B3x%2F%28x-5%29=-1%2F%28x-3%29
This is my work: 

First off, I marked that x cannot equal 5 or 3, because of how(x-3) and (x-5) are in the denominator and 
I can't have 0 in the denominator. 

I turned it into %287x%2B1%29%2F%28x-5%29%28x-3%29%2B3x%2F%28x-5%29=-1%28x-3%29    ( <<<--- actually, right side is -1%2F%28x-3%29, but I consider your error as a typo here )

I noticed that after I factored out x%5E2-8x%2B15 I could rationalize the denominator to (x-5)(x-3) by multiplying 3x/(x-5) 

with (x-3) and -1/(x-3) with (x-5) and cancel it all out to just end up with %287x%2B1%29%2B3x%28x-3%29=-1%28x-5%29.           <<<--- CORRECT!

Next, I simplified it down to 7x%2B1%2B3x%5E2-9x=-x%2B5 to 3x%5E2-x-4=0.                      <<<--- CORRECT!

I didn't know if I could factor it out anymore mentally, so I chose my safest bet which was using 
the quadratic formula to get x=4/3 and x=-1.                                                                     <<<--- CORRECT!

Those two answers would work because it's not 5 or 3 (because then it would equal 0).                            <<<--- CORRECT!

Am I right? Or did I do something wrong? I hope I was clear enough and I would really appreciate the help!    <<<---  You are right. No error. At least, I do not see it.