Question

How do you simplify `((r+s)/(x^2-y^2))/((r+s)^2/(x-y))`?

Answer 1

See Below.

Explanation:

We have,

`((r + s)/(x^2 - y^2))/((r + s)^2/(x - y)) = ((r + s)/((x + y)(x - y)))/(((r + s)(r + s))/((x - y))`

`=cancel((r +s))/((x + y)cancel((x - y))) xx cancel((x - y))/(cancel((r+ s))(r + s))` [As `a/b = a xx 1/b`]

`= 1/((r + s)(x + y))`

`= 1/(rx + sx + ry + sy)`

Hope this helps.

Answer 2

`1/((x+y)(r+s))`

Explanation:

break it up like this:

`(r+s)/(x^2-y^2)div(r+s)^2/(x-y)`

when dividing remember to keep the dividend, then change the division sign to a multiplication sign, and then use the reciprocal of the divisor.

`(r+s)/(x^2-y^2)*(x-y)/(r+s)^2`

you can simplify the expression further. You can factor `x^2-y^2` to `(x+y)(x-y)`

`(r+s)/((x+y)(x-y))*(x-y)/((r+s)(r+s))`

now just multiply the two of these like you would do for normal fractions

`(cancel((r+s))cancel((x-y)))/((x+y)cancel((x-y))cancel((r+s))(r+s))`

some of the terms are going to cancel each other out leaving you with this answer
`1/((x+y)(r+s))`