Question

Using the properties of proportion, find the value of $\frac {p+2x}{p-2x} + \frac {p+2y}{p-2y} , if p = \frac {4xy}{x+y}$

Answer 1

Please see the attachment

Answer 2

Solution:

if $p = \frac {4xy}{x+y}$

then to solve the given expression,put the value of p in it

$\frac {p+2x}{p-2x} + \frac {p+2y}{p-2y} \\ \\ =\frac { \frac{4xy}{x + y} +2x}{ \frac{4xy}{x + y} -2x} + \frac { \frac{4xy}{x + y} +2y}{ \frac{4xy}{x + y} -2y} \\ \\ = \frac { 2x(\frac{2y}{x + y} +1)}{ 2x(\frac{2y}{x + y} -1)} + \frac {2y( \frac{2x}{x + y} +1)}{ 2y(\frac{2x}{x + y} -1)} \\ \\ = \frac { (\frac{2y + x + y}{x + y})}{ (\frac{2y - x - y}{x + y})} + \frac {( \frac{2x + x + y}{x + y})}{ (\frac{2x - x - y}{x + y})} \\ \\ = \frac{3y + x}{y - x} + \frac{3x + y}{x - y} \\ \\ = \frac{3y + x}{y - x} - \frac{3x + y}{y - x} \\ \\ = \frac{(3y + x) - (3x + y)}{y - x} \\ \\ = \frac{3y - y + x - 3x}{y - x} \\ \\ = \frac{2y - 2x}{y - x} \\ \\ = \frac{2(y - x)}{y - x} \\ \\ = 2$
Hope it helps you.