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

May 24, 2015

Given: $\left[\textcolor{red}{\frac{2 x + y}{x - y}}\right] - \left[\textcolor{b l u e}{\frac{3 x - y}{x + y}}\right] - \left[\textcolor{g r e e n}{5 \frac{x + 2 y}{{y}^{2} - {x}^{2}}}\right]$

Step 1: determine an appropriate common denominator
Since
$\left({x}^{2} - {y}^{2}\right)$
$= \left(x + y\right) \cdot \textcolor{red}{\left(x - y\right)}$
$= \left(x - y\right) \cdot \textcolor{b l u e}{\left(x + y\right)}$
$= \left(- 1\right) \cdot \textcolor{g r e e n}{\left({y}^{2} - {x}^{2}\right)}$
${x}^{2} - {y}^{2}$ 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
$+ \textcolor{red}{\left(2 x + y\right)} \cdot \left(x + y\right) = + \left(2 {x}^{2} + 3 x y + {y}^{2}\right)$
$- \textcolor{b l u e}{\left(3 x - y\right)} \cdot \left(x - y\right) = - \left(3 {x}^{2} - 4 x y + {y}^{2}\right)$
$- \textcolor{g r e e n}{\left(5 x + 2 y\right)} \cdot \left(- 1\right) = + \left(5 x + 2 y\right)$

Giving a numerator sum of
$- {x}^{2} + 7 x y + 5 x + 2 y$

Step 3: Recombine as a final solution
$\frac{- {x}^{2} + 7 x y + 5 x + 2 y}{{x}^{2} - {y}^{2}}$

May 24, 2015

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 $\left(a - b\right) \left(a + b\right)$. Thus,

${y}^{2} - {x}^{2} = \left(y - x\right) \left(y + x\right)$

We can see that this fits as l.c.d for the second fraction's denominator (due to the term $\left(x + y\right)$) 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 $\left(y - x\right)$, thus $\left(y - x\right) = \left(- 1\right) \left(x - y\right)$

So, our l.c.d. will be $\left(- 1\right) \left(x - y\right) \left(x + y\right)$

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)

$\frac{\left(- 1\right) \left(x + y\right) \left(2 x + y\right) - \left(- 1\right) \left(x - y\right) \left(3 x - y\right) - \left(5 x + 2 y\right)}{\left(- 1\right) \left(x - y\right) \left(x + y\right)}$

Aggregating and simplifying signals and factors:

$\frac{- \left(x + y\right) \left(2 x + y\right) + \left(x - y\right) \left(3 x - y\right) - 5 x - 2 y}{{y}^{2} - {x}^{2}}$

$\frac{2 {x}^{2} + 3 x y + {y}^{2} + 3 {x}^{2} - 4 x y + {y}^{2} - 5 x - 2 y}{{y}^{2} - {x}^{2}}$

$\frac{5 {x}^{2} - 5 x - x y + {y}^{2} - 2 y}{{y}^{2} - {x}^{2}}$

$\textcolor{g r e e n}{\frac{5 x \left(x - 1\right) + y \left(y - x - 2\right)}{{y}^{2} - {x}^{2}}}$