Question

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

A
0
B
1
C
1
D
2

Solution

## 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 $$\frac{a+2x}{a-2x}+\frac{a+2y}{a-2y} = \frac{x+3y}{-x+y}+\frac{3x+2y}{x-y}$$$$= \frac{x^{2}-xy+3xy-3y^{2}-3x^{2}-2xy+3xy+y^{2}}{-x^{2}+2xy-y^{2}}$$$$= \frac{-2x^{2}-2y^{2}+4xy}{-x^{2}+2xy-y^{2}} = \frac{-2[x^{2}+y^{2}-2xy]}{-[x^{2}+y^{2}-2xy]}$$$$= 2$$ Mathematics

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