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

2 Answers
Jan 16, 2017

Answer:

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.

Jan 16, 2017

Answer:

#(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#