# If p=4xy/(x+y) then how much is (p+2x)/(p−2x)+(p+2y)/(p−2y)?

Jan 16, 2017

If $p = 4 x \frac{y}{x + y}$
then
$\textcolor{w h i t e}{\text{XXX}} \frac{p + 2 x}{p - 2 x} + \frac{p + 2 y}{p - 2 y} = \textcolor{g r e e n}{2}$

#### Explanation:

There may be a simpler way to get this result, but I am not certain that I have interpreted the question correctly, so I will just use a sledgehammer approach.

Note 1: I've used $\frac{4 x y}{x + y}$ in place of the form given: $4 x \frac{y}{x + y}$
but these are equivalent.

Note 2: I've also assumed that $x \ne y$ and neither $x$ not $y$ are $= 0$

Part 1: Simplifying $\frac{p + 2 x}{p - 2 x}$

$\frac{p + 2 x}{p - 2 x}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{\frac{4 x y}{x + y} + 2 x}{\frac{4 x y}{x + y} - 2 x}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\frac{4 x y + 2 {x}^{2} + 2 x y}{x + y}}{\frac{4 x y - 2 {x}^{2} - 2 x y}{x + y}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{4 x y + 2 {x}^{2} + 2 x y}{4 x y - 2 {x}^{2} - 2 x y}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{6 x y + 2 {x}^{2}}{2 x y - 2 {x}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{3 y + x}{y - x}$

Part 2: Simplifying $\frac{p + 2 y}{p - 2 y}$
Omitting the details but following the same process as above:
$\frac{p + 2 y}{p - 2 y}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{3 x + y}{x - y}$

Part 3: Evaluating the Sum $\frac{p + 2 x}{p - 2 x} + \frac{p + 2 y}{p - 2 y}$
$\frac{p + 2 x}{p - 2 x} + \frac{p + 2 y}{p - 2 y}$
$\textcolor{w h i t e}{\text{XXX}} = \frac{3 y + x}{y - x} + \frac{3 x + y}{x - y}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\left(3 y + x\right) - \left(3 x + y\right)}{y - x}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2 y - 2 x}{y - x}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{2 \left(\cancel{y - x}\right)}{\cancel{y - x}}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The assumption that $x + y \ne 0$ is implied by the definition of $p$
but the effect of the other assumptions: x!=y; x!=0; y!=0
should be evaluated.

Jan 16, 2017

$\frac{p + 2 x}{p - 2 x} + \frac{p + 2 y}{p - 2 y} = 2$

#### Explanation:

Let us use here the concept of Componendo and Dividendo. This essentially states that if $\frac{r}{s} = \frac{u}{v}$, then $\frac{r + s}{r - s} = \frac{u + v}{u - v}$.

As $\frac{r}{s} = \frac{u}{v}$, adding $1$ to each side we get $\frac{r}{s} + 1 = \frac{u}{v} + 1$ or $\frac{r + s}{s} = \frac{u + v}{v}$. This is called Componendo.

and subtracting $1$ from each side we get $\frac{r}{s} - 1 = \frac{u}{v} - 1$ or $\frac{r - s}{s} = \frac{u - v}{v}$. This is called Dividendo.

Their ratio gives us $\frac{r + s}{r - s} = \frac{u + v}{u - v}$, which is called applying Componendo and Dividendo together.

Coming to the problem $p = 4 x \frac{y}{x + y} = \frac{4 x y}{x + y}$

i.e. $\frac{p}{2 x} = \frac{2 y}{x + y}$

Now applying Componendo and Dividendo, we get

$\frac{p + 2 x}{p - 2 x} = \frac{2 y + x + y}{2 y - x - y}$ ...............(A)

and as $\frac{p}{2 y} = \frac{2 x}{x + y}$

applying Componendo and Dividendo, we get

$\frac{p + 2 y}{p - 2 y} = \frac{2 x + x + y}{2 x - x - y}$ ...............(B)

Adding (A) and (B) , we get

$\frac{p + 2 y}{p - 2 y} + \frac{p + 2 y}{p - 2 y} = \frac{2 y + x + y}{2 y - x - y} + \frac{2 x + x + y}{2 x - x - y}$

$= \frac{3 y + x}{y - x} + \frac{3 x + y}{x - y}$

$= \frac{- 3 y - x + 3 x + y}{x - y}$

$= \frac{2 x - 2 y}{x - y} = 2$