Question

if p=4xy/x+y,find the value of p+2x/p-2x+p+2y/p-2y

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

2.

Step-by-step explanation:

Given:- p = 4xy/( x + y )

To Find:- Value of $\frac{p + 2x}{p - 2x} + \frac{p+2y}{p-2y}$.

Solution:-

It is given that p = 4xy/( x + y ) ------ ( 1 )

Dividing equation ( 1 ) by 2x on both the sides of the equation, we get

$\frac{p}{2x}$ = $\frac{4xy}{(x+y)2x}$

⇒ $\frac{p}{2x}$ = $\frac{2y}{(x+y)}$

Applying Componendo and Dividendo on the above equation, we get

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

⇒ $\frac{p+2x}{p-2x}$ = $\frac{3y+x}{y-x}$     ----------- ( 2 )

Dividing equation ( 1 ) by 2y on both the sides of the equation, we get

$\frac{p}{2y}$ = $\frac{4xy}{(x+y)2y}$

⇒ $\frac{p}{2y}$ = $\frac{2x}{(x+y)}$

Again applying Componendo and Dividendo on the above equation, we get

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

⇒ $\frac{p+2y}{p-2y}$ = $\frac{3x+y}{x-y}$     ------------ ( 3 )

Now on doing( 2 ) + ( 3 ), we get

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

⇒ $\frac{p+2x}{p-2x} + \frac{p+2y}{p-2y}$ = $\frac{3x+y}{x-y} - \frac{(3y+x)}{x-y}$

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

⇒ $\frac{p+2x}{p-2x} + \frac{p+2y}{p-2y}$ = $\frac{2x-2y}{x-y}$

⇒ $\frac{p+2x}{p-2x} + \frac{p+2y}{p-2y}$ = 2

Therefore, the value of $\frac{p+2x}{p-2x} + \frac{p+2y}{p-2y}$ is 2.

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