Verified answer

x^2+2xy+y^2 to get (x+y)(x+y)

(x+y)/(x+y)(x+y)

Cancel out (x+y)

And you get 1/(x+y)

if you reversed the two terms on the division symbol, then it would be (x+y) ... so that makes it 1/(x+y). this logic comes from x^2+2xy+y^2=(x+y)^2 or (x+y)*(x+y) <--pascal's triangle ... there is more than one method for solving a problem like this. You could intuitively change x^2+2xy+y^2 to (x+y)*(x+y) and then reduce the fraction or you could set it up like a long division problem. the first term would be x ... which gives you x^2+xy ... the remainder is xy ... subtraction leaves you with xy ... so add y to the top ... y times the first term x gives you xy + (y times the second term) y^2 ... subtraction =0.

(x+y)/(x^2+2xy+y^2)

I'll work with just the bottom right now OK?

(x^2+2xy+y^2) = (x + y)(x + y) [FOIL that out, you can see it checks out.]

Now lets put that back in the original problem.

(x+y)/(x+y)(x+y) -> (You will see one of the (x+y)'s cancel out.)

After cancelling you get your final answer of 1/(x+y).

Enjoy ! :)

First factorise x^2+2xy+y^2 to get (x+y)(x+y)

(x+y)/(x+y)(x+y)

Cancel out (x+y)

And you get 1/(x+y)

(x + y)/(x^2 + 2xy + y^2) = (x + y)/(x + y)² = 1/(x + y)

= (x+y) ÷ (x+y) (x+y)

then cancel (x+y) and (x+y)

= 1 ÷ (x+y)