Question

If $a=\frac{4xy}{x+y},$ the value of $\frac{a+2x}{a-2x}+\frac{a+2y}{a-2y}$ In most simplified form is

• A. $0$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is D $2$

Solution : $a=\frac{4xy}{x+y}$

then $\frac{a+2x}{a-2x}=\frac{\frac{4xy}{x+y}+2x}{\frac{4xy}{x+y}-2x}=\frac{4xy+2x^2+2xy}{4xy-2x^2-2xy}=\frac{2x^3+6xy}{-2x^2+2xy}$

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

And $\frac{a+2y}{a-2y}=\frac{\frac{4xy}{x+y}+2y}{\frac{4xy}{x+y}-2y}=\frac{4xy+2xy+2y^2}{2xy-2y^2}=\frac{6xy+2y^2}{2xy-2y^2}$

$=\frac{3x+2y}{x-y}$

then $\frac{a+2x}{a-2x}+\frac{a+2y}{a-2y}=\frac{x+3y}{-x+y}+\frac{3x+2y}{x-y}$

$=\frac{x^2-xy+3xy-3y^2-3x^2-2xy+3xy+y^2}{-x^2+2xy-y^2}$

$=\frac{-2x^2-2y^2+4xy}{-x^2+2xy-y^2}=\frac{-2[x^2+y^2-2xy]}{-[x^2+y^2-2xy]}$

$=2$