SOLUTION: If I have a problem like this one I am not sure how to even start to simplify it. The directions say Perfom the indicated operations and simplify: x+3y x+

Algebra ->  Expressions-with-variables -> SOLUTION: If I have a problem like this one I am not sure how to even start to simplify it. The directions say Perfom the indicated operations and simplify: x+3y x+      Log On


   

Question 597185: If I have a problem like this one I am not sure how to even start to simplify it. The directions say Perfom the indicated operations and simplify:
x+3y x+y x+2y
_______________ - ___________________ + _____________
4(x^2)+12xy+8(y^2) 4(x^2)+20xy+24(y^2) 4(x^2)+16xy+12(y^2)

I tried to take ouf a 4 of each denominator then got stuck, can you help? the numerators are x+3y, x+y, and x+2y, they wont let me put them right, sorry
Found 2 solutions by stanbon, KMST:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
x+3y x+y x+2y
_______________ - ___________________ + _____________
4(x^2)+12xy+8(y^2) 4(x^2)+20xy+24(y^2) 4(x^2)+16xy+12(y^2)

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You need a least common denominator to and and/or subtract fractions.
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You need to factor each of the given denominators and determine
the lcm.
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Then rewrite each fraction with the lcm as its denominator.
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Then combine the numerators over the lcm.
----
Cheers,
Stan H.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You are right in thinking that you need to do some factoring and taking out a 4 from each denominator is a good start.
I'll show you the next steps, and what clues and tricks I used.

THE SOLUTION (skipping explanations for now):

=

=

=

=

=

=%28x%5E2%2B8xy%2B12y%5E2%29%2F%284%28x%2By%29%28x%2B2y%29%28x%2B3y%29%29

=%28%28x%2B2y%29%28x%2B6y%29%29%2F%284%28x%2By%29%28x%2B2y%29%28x%2B3y%29%29

=%28x%2B6y%29%2F%284%28x%2By%29%28x%2B3y%29%29

ABOUT FACTORING:
When you have a polynomial with x and y, like
x%5E2%2B3xy%2B2y%5E2
factoring may look complicated, but there are clues and tricks that helped me factor it.
I noticed that he total degree (in x and y) of all the terms is 2, so I knew that

and renaming variables using z=x%2Fy the big bracket turns into z%5E2%2B3z%2B2
I knew I could handle that.
It is easier to factor when there is only one variable, and if it seemed difficult I could always resort to the quadratic formula to find that the zeros of that polynomial. It would tell me that the zeros are z=-1 and z=-2, and from there I would figure out that the factors are (z+1) and (z+2).