factoring (x+3y)/(x^2+2xy+2y) + (x-y)/(x^2+4xy+3y^2)

Question

I already know the is 2(x^2+3xy+4y^2)/(x+3y)(x+y)^2 I just want to know the steps that were taken to get this answer

Answer 1

I think you misstated the question.

(x+3y)/(x²+2xy+y²)+(x-y)/(x²+4xy+3y²).

Note that x²+2xy+y²=(x+y)² and x²+4xy+3y²=(x+3y)(x+y).

The LCD of the denominators is (x+3y)(x+y)².

The expression becomes:

[(x+3y)(x+3y)+(x-y)(x+y)]/[(x+3y)(x+y)²]=

(x²+6xy+9y²+x²-y²)/[(x+3y)(x+y)²]=

(2x²+6xy+8y²)/[(x+3y)(x+y)²]=

2(x²+3xy+4y²)/[(x+3y)(x+y)²].