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

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Answer 1 · 2018-05-29T04:48:56

See Below.

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 · 2018-05-29T05:15:36

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

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

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