Question

How do you combine `[(2x + y)/(x - y)] - [(3x - y)/(x + y)] - [(5x + 2y)/(y^2 - x^2)]`?

Answer 1

Given: `[color(red)((2x+y)/(x-y))] - [color(blue)((3x-y)/(x+y))] -[color(green)(5(x+2y)/(y^2-x^2))]`

Step 1: determine an appropriate common denominator
Since
`(x^2-y^2)`
`= (x+y) * color(red)((x-y))`
`=(x-y) * color(blue)((x+y))`
`=(-1) * color(green)((y^2-x^2))`
`x^2-y^2` would seem to be the obvious choice for a common denominator

Step 2: Evaluate the numerators by multiplying each by the factor needed to obtain the common denominator
`([color(red)((2x+y)) * (x+y)] - [color(blue)((3x-y)) * (x-y)] -[color(green)((5x+2y)) * (-1)])/((x^2-y^2)`

It is probably easiest to evaluate each numerator term separately then recombine
`+color(red)((2x+y)) * (x+y) = +(2x^2+3xy+y^2)`
`-color(blue)((3x-y)) * (x-y) = -(3x^2-4xy+y^2)`
`-color(green)((5x+2y)) * (-1) = +(5x+2y)`

Giving a numerator sum of
`-x^2+7xy+5x+2y`

Step 3: Recombine as a final solution
`(-x^2+7xy+5x+2y)/(x^2-y^2)`

Answer 2

We can find the lowest common denominator for your fractions. In order to do that, we'd better factor the third fraction's denominator, as it's a squared product and the others have degree one (`x^1`).

By factoring laws, we know that if `a^2-b^2` is the result, then it comes from `(a-b)(a+b)`. Thus,

`y^2-x^2=(y-x)(y+x)`

We can see that this fits as l.c.d for the second fraction's denominator (due to the term `(x+y)`) but it doesn't really fit the first one's. However, we can see that the denominator of the first one is the very opposite of `(y-x)`, thus `(y-x)=(-1)(x-y)`

So, our l.c.d. will be `(-1)(x-y)(x+y)`

Now, rewriting the third fraction and then proceeding to the calculation:

`(2x+y)/(x-y)-(3x-y)/(x+y)-(5x+2y)/((-1)(x-y)(x+y)`

`((-1)(x+y)(2x+y)-(-1)(x-y)(3x-y)-(5x+2y))/((-1)(x-y)(x+y))`

Aggregating and simplifying signals and factors:

`(-(x+y)(2x+y)+(x-y)(3x-y)-5x-2y)/(y^2-x^2)`

`(2x^2+3xy+y^2+3x^2-4xy+y^2-5x-2y)/(y^2-x^2)`

`(5x^2-5x-xy+y^2-2y)/(y^2-x^2)`

`color(green)((5x(x-1)+y(y-x-2))/(y^2-x^2))`