SOLUTION: ((x)/(x-4)-(2)/(x+3))/((x^2-4x)/(2x+6)) These problems are giving me a hard time I know you have to get a common denominator and everything but for some reason i get stuck.

Algebra ->  Rational-functions -> SOLUTION: ((x)/(x-4)-(2)/(x+3))/((x^2-4x)/(2x+6)) These problems are giving me a hard time I know you have to get a common denominator and everything but for some reason i get stuck.      Log On


   

Question 22911: ((x)/(x-4)-(2)/(x+3))/((x^2-4x)/(2x+6))
These problems are giving me a hard time I know you have to get a common denominator and everything but for some reason i get stuck.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, congratulations on getting all the parentheses in the right place in typing this problem. If you put 3 sets of set braces before and 3 after this, then your question will be expressed in perfect form for a complex fraction:
+%28%28x%29%2F%28x-4%29-%282%29%2F%28x%2B3%29%29%2F%28%28x%5E2-4x%29%2F%282x%2B6%29%29

Let's just "unstack" the problem and write it like this:
+%28%28x%29%2F%28x-4%29-%282%29%2F%28x%2B3%29%29 divided by %28%28x%5E2-4x%29%2F%282x%2B6%29%29

On the left side, you need to find a common denominator, which is (x-4)(x+3). Multiply the first fraction by %28x%2B3%29%2F%28x%2B3%29, and the second fraction by %28x-4%29%2F%28X-4%29. It should look like this:
divided by %28%28x%5E2-4x%29%2F%282x%2B6%29%29

Now, put down the common denominator for the first fraction; combine numerators to form the first numerator, and invert and factor the second fraction:
+%28x%5E2%2B3x+-2x%2B8%29%2F%28%28x-4%29%28x%2B3%29%29 times %282%28x%2B3%29%29%2F+%28x%28x-4%29%29++

Combine like terms in the first numerator, divide out the (x+3) factors in the first denominator and the second numerator:
+%28x%5E2%2Bx+%2B8%29%2F%28x-4%29 times 2%2F+%28x%28x-4%29%29++

This brings (I hope!!) the final answer:
+%282%2A%28x%5E2%2Bx+%2B8%29%29%2F%28x%2A%28x-4%29%5E2%29

R^2 at SCC

P.S. That was ugly!! Somebody let me know if there might be an error in all of that!!