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 assume given expression  is (x+3y)/(x^2+2xy+y^2) +(x-y)/(x^2+4xy+3y^2) ----> (1)

Now (x+3y)/(x^2+2xy+y^2)

= (x+3y)/(x+y)^2

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

Now the denominator of above fraction x^2+4xy+3y^2

Factorize the multistep expression x^2+4xy+3y^2 .

x^2+3xy+xy+3y^2

= x(x+3y)+y(x+3y)

= (x+y)(x+3y)

Plug the values in (1).

= (x+3y)/(x+y)^2 +(x-y)/(x+y)(x+3y)

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

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

= (x^2+9y^2+6xy+x^2-y^2]/(x+3y)(x+y)^2

= (2x^2+6xy+8y^2)/(x+3y)(x+y)^2

= 2(x^2+3xy+4y)/(x+3y)(x+y)^2