Question

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

Answer 1

If `p=4xy/(x+y)`
then
`color(white)("XXX")(p+2x)/(p-2x)+(p+2y)/(p-2y)=color(green)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 `(4xy)/(x+y)` in place of the form given: `4xy/(x+y)`
but these are equivalent.

Note 2: I've also assumed that `x!=y` and neither `x` not `y` are `=0`

Part 1: Simplifying `(p+2x)/(p-2x)`

`(p+2x)/(p-2x)`
`color(white)("XXX")=((4xy)/(x+y)+2x)/((4xy)/(x+y)-2x)`

`color(white)("XXX")=((4xy+2x^2+2xy)/(x+y))/((4xy-2x^2-2xy)/(x+y))`

`color(white)("XXX")=(4xy+2x^2+2xy)/(4xy-2x^2-2xy)`

`color(white)("XXX")=(6xy+2x^2)/(2xy-2x^2)`

`color(white)("XXX")=(3y+x)/(y-x)`

Part 2: Simplifying `(p+2y)/(p-2y)`
Omitting the details but following the same process as above:
`(p+2y)/(p-2y)`
`color(white)("XXX")=(3x+y)/(x-y)`

Part 3: Evaluating the Sum `(p+2x)/(p-2x)+(p+2y)/(p-2y)`
`(p+2x)/(p-2x)+(p+2y)/(p-2y)`
`color(white)("XXX")=(3y+x)/(y-x)+(3x+y)/(x-y)`

`color(white)("XXX")=((3y+x)-(3x+y))/(y-x)`

`color(white)("XXX")=(2y-2x)/(y-x)`

`color(white)("XXX")=(2 (cancel(y-x)))/cancel(y-x)`

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

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

Answer 2

`(p+2x)/(p-2x)+(p+2y)/(p-2y)=2`

Explanation:

Let us use here the concept of Componendo and Dividendo. This essentially states that if `r/s=u/v`, then `(r+s)/(r-s)=(u+v)/(u-v)`.

As `r/s=u/v`, adding `1` to each side we get `r/s+1=u/v+1` or `(r+s)/s=(u+v)/v`. This is called Componendo.

and subtracting `1` from each side we get `r/s-1=u/v-1` or `(r-s)/s=(u-v)/v`. This is called Dividendo.

Their ratio gives us `(r+s)/(r-s)=(u+v)/(u-v)`, which is called applying Componendo and Dividendo together.

Coming to the problem `p=4xy/(x+y)=(4xy)/(x+y)`

i.e. `p/(2x)=(2y)/(x+y)`

Now applying Componendo and Dividendo, we get

`(p+2x)/(p-2x)=(2y+x+y)/(2y-x-y)` ...............(A)

and as `p/(2y)=(2x)/(x+y)`

applying Componendo and Dividendo, we get

`(p+2y)/(p-2y)=(2x+x+y)/(2x-x-y)` ...............(B)

Adding (A) and (B) , we get

`(p+2y)/(p-2y)+(p+2y)/(p-2y)=(2y+x+y)/(2y-x-y)+(2x+x+y)/(2x-x-y)`

`=(3y+x)/(y-x)+(3x+y)/(x-y)`

`=(-3y-x+3x+y)/(x-y)`

`=(2x-2y)/(x-y)=2`