Question

Simplify this expression. (2/x-y) + (3/x+y) - (5/y+x) - (7x-9y/y^2-x^2) = ___________/ y^2 - x^2

Answer 1

`\frac{ - 17x + 5y}{ {y}^(2) - {x}^(2) }`

Solution,

`(2)/(x - y) + (3)/(x + y) - (5)/(y - x) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (2)/(x - y) + (3)/(x + y) - (5)/( - (x - y)) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (2)/(x - y) + (3)/(x + y) + (5)/(x - y) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (2)/(x - y) + (5)/(x - y) + (3)/(x + y) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (2 + 5)/(x - y) + (3)/(x + y) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (7)/(x - y) + (3)/(x + y) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = (7(x + y) + 3(x - y))/((x - y)(x + y)) - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{7x + 7y + 3x - 3y}{ {x}^(2) - {y}^(2) } - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{10x + 4y}{ {x}^(2) - {y}^(2) } - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{10x + 4y}{ - ( {y}^(2) - {x)}^(2) } - \frac{7x - 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{ - 10x - 4y - 7x + 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{ - 10x - 7x - 4y + 9y}{ {y}^(2) - {x}^(2) } \\ = \frac{ - 17x + 5y}{ {y}^(2) - {x}^(2) }`

hope this helps...

Good luck on your assignment...

Answer 2

(5y-17x)/ (y^2-x^2)

Explanation:

(2/x-y) + (3/x+y) - (5/y-x) - (7x-9y/y^2-x^2) =

Get a common denominator of y^2 - x^2

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

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

-5/(y-x) = -5/(y-x) *(y+x)/(y+x) = -5y-5x / (y^2 -x^2)

- (7x-9y/(y^2-x^2) = -7x +9y/ (y^2-x^2)

Combine the numerators since the denominators are equal

-2y -2x +3y-3x-5y-5x-7x+9y

5y-17x

Put this over the denominator

(5y-17x)/ (y^2-x^2)