SOLUTION: I am asked to solve {{{(k)/(2k+2)}}}+{{{(2k-16)/ (4k+4)}}}={{{(k-3)/(k+1)}}} Thank you.

Algebra ->  Equations -> SOLUTION: I am asked to solve {{{(k)/(2k+2)}}}+{{{(2k-16)/ (4k+4)}}}={{{(k-3)/(k+1)}}} Thank you.      Log On


   

Question 236582: I am asked to solve
%28k%29%2F%282k%2B2%29+%282k-16%29%2F+%284k%2B4%29=%28k-3%29%2F%28k%2B1%29
Thank you.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

%28k%29%2F%282k%2B2%29+%282k-16%29%2F+%284k%2B4%29=%28k-3%29%2F%28k%2B1%29

Factor everything that will factor:

%28k%29%2F%282%28k%2B1%29%29+%282%28k-8%29%29%2F+%284%28k%2B1%29%29=%28k-3%29%2F%28k%2B1%29

%28k%29%2F%282%28k%2B1%29%29+%28cross%282%29%28k-8%29%29%2F+%28cross%284%29%28k%2B1%29%29=%28k-3%29%2F%28k%2B1%29
          2

%28k%29%2F%282%28k%2B1%29%29+%28k-8%29%2F%282%28k%2B1%29%29=%28k-3%29%2F%28k%2B1%29

LCD = 2%28k%2B1%29

Multiply through by that over 1, %282%28k%2B1%29%29%2F1

%28k%29%2F%282%28k%2B1%29%29%282%28k%2B1%29%29%2F1+%28k-8%29%2F%282%28k%2B1%29%29%282%28k%2B1%29%29%2F1=%28k-3%29%2F%28k%2B1%29%282%28k%2B1%29%29%2F1

%28k%29%2F%28cross%282%29%28cross%28k%2B1%29%29%29%28cross%282%29%28cross%28k%2B1%29%29%29%2F1+%28k-8%29%2F%28cross%282%29%28cross%28k%2B1%29%29%29%28cross%282%29%28cross%28k%2B1%29%29%29%2F1=%28k-3%29%2F%28cross%28k%2B1%29%29%282%28cross%28k%2B1%29%29%29%2F1

k+%2B+k-8+=+2%28k-3%29

2k-8=2k-6

-8=-6

That is a false numerical equation, so there is no solution.

Edwin