SOLUTION: x+4/x+2 + 32/x^2+9x+14=3 my work: x+4/x+2 + 32/(x+2)(x+7)=3 (x+2)(x+7)(x+4/x+2 +32/(x+2(x+7)=3(x+2)(x+7) What's next? Am I doing this right?

Algebra ->  Equations -> SOLUTION: x+4/x+2 + 32/x^2+9x+14=3 my work: x+4/x+2 + 32/(x+2)(x+7)=3 (x+2)(x+7)(x+4/x+2 +32/(x+2(x+7)=3(x+2)(x+7) What's next? Am I doing this right?      Log On


   

Question 193232This question is from textbook saxon algebra 2
: x+4/x+2 + 32/x^2+9x+14=3
my work: x+4/x+2 + 32/(x+2)(x+7)=3
(x+2)(x+7)(x+4/x+2 +32/(x+2(x+7)=3(x+2)(x+7)
What's next?
Am I doing this right? This question is from textbook saxon algebra 2

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B4%29%2F%28x%2B2%29 + 32%2F%28x%5E2%2B9x%2B14%29 = 3
:
my work:
%28x%2B4%29%2F%28x%2B2%29 + 32%2F%28%28x%2B2%29%28x%2B7%29%29 = 3
:
Multiply by the common denominator
(x+2)(x+7)*%28x%2B4%29%2F%28x%2B2%29 + (x+2)(x+7)*32%2F%28%28x%2B2%29%28x%2B7%29%29 = 3(x+2)(x+7)
Next, Cancel out the denominators and you have:
(x+7)(x+4) + 32 = 3(x^2+9x+ 14)
:
x^2 + 11x + 28 + 32 = 3x^2 + 27x + 42
:
x^2 + 11x + 60 = 3x^2 + 27x + 42
:
arrange as a quadratic equation on the right
0 = 3x^2 - x^2 + 27x - 11x + 42 - 60
:
0 = 2x^2 + 16x - 18
Factor to:
(x+9)(2x-2) = 0
Two solutions:
x = -9
and
2x = 2
x = 1
:
:
Check solutions in original problem
x = -9
%28-9%2B4%29%2F%28-9%2B2%29 + 32%2F%2881-81%2B14%29 = 3
%28-5%29%2F%28-7%29 + 32%2F14 = 3
10%2F14 + 32%2F14 = 3
42%2F14 = 3
:
you can check the solution; x=1