Simplify [1+y/(x+y)]/ [1-3y2/(x2-y2)]

Question

a.x/y-x
b.y/x-y
c.x-y/2-2x
d.x-y/x-2y

Answer 1

1 + 7/(x+y) = (x+2y)/(x+y)
1 - 3y^2/(x^2-y^2) = (x^2-4y^2)/(x^2-y^2)
so, the result is

(x+2y)/(x+y) * (x+y)(x-y) / (x-2y)(x+2y) = (x-y)/(x-2y)

Answer 2

i get the answer but ,,still i wonder how you get that equation ,, when i try to solve it,this is the output
(1+y/x+y)/(x^2-y^2/1-3y^2).can you explain it to me further?how u get that equation above?

tnx mr. steve

Answer 3

Huh? I clearly showed the factors of each polynomial. When you divide by a fraction, it is the same as multiplying by its reciprocal.

If you can't follow those steps, I fear you have a long road ahead...

Sorry about the typo. 7 is near the y on the keyboard. I meant

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

Suppose you had

1 + 3/8

you put everything over a common denominator if 8, so you have

8/8 + 3/8 = (8+3)/8

1 + y/(x+y) is done the same way

(x+y)/(x+y) + y/(x+y)
(x+y+y)/(x+y)
(x+2y)/(x+y)