How do you simplify $\frac{2 x + 7}{x^{2} - y^{2}} + \frac{- 5}{x^{2} - 2 x y + y^{2}}$ $(2x+7)/(x^2-y^2)+(-5)/(x^2-2xy+y^2)$ ?

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Answer 1 · 2018-05-26T08:23:08

see below

$\frac{2 x + 7}{x^{2} - y^{2}} + \frac{- 5}{x^{2} - 2 x y + y^{2}}$ $(2x+7)/(x^2-y^2)+ (-5)/(x^2-2xy+y^2)$

firstly factorise the denominators

$\frac{2 x + 7}{(x + y) (x - y)} - \frac{5}{{(x - y)}^{2}}$ $(2x+7)/((x+y)(x-y))-5/(x-y)^2$

we need the denominators the same

multiply top /bottom of first fraction by $(x - y)$ $(x-y)$

multiply top/bottom of second fraction by $(x + y)$ $(x+y)$

$\frac{(2 x + 7) (x - y)}{(x + y) {(x - y)}^{2}} - \frac{5 (x + y)}{(x + y) {(x - y)}^{2}}$ $((2x+7)(x-y))/((x+y)(x-y)^2)-(5(x+y))/((x+y)(x-y)^2)$

now simplify

$\frac{2 x^{2} - 2 x y + 7 x - 7 y - 5 x - 5 y}{(x + y) {(x - y)}^{2}}$ $(2x^2-2xy+7x-7y-5x-5y)/((x+y)(x-y)^2)$

$\frac{2 x^{2} - 2 x y + 2 x - 12 y}{(x + y) {(x - y)}^{2}}$ $(2x^2-2xy+2x-12y)/((x+y)(x-y)^2)$

$\frac{2 (x^{2} - x y + x - 6 y)}{(x + y) {(x - y)}^{2}}$ $(2(x^2-xy+x-6y))/((x+y)(x-y)^2)$

How do you simplify (2x+7)/(x^2-y^2)+(-5)/(x^2-2xy+y^2)2x+7x2−y2+−5x2−2xy+y2(2x+7)/(x^2-y^2)+(-5)/(x^2-2xy+y^2)?