Question

please answer this question​

Answer 1

$\bold{\underline{\underline{\underbrace{Answer}}}}$

x = p and y = -q

Step-by-step explanation:

• To find

The value of x and y

• Given

px + qy = p² - q²(equation 1)

qx - py = 2py

• Solution

{Taking equation 2}

$\sf qx - py = 2py$

{Taking py on RHS}

$\sf qx = 2py \ + py$

{Dividing both sides by q

$\sf x = \frac{2pq}{q} + \frac{py}{q} \\ \\ \sf x = 2p + \frac{py}{q} (equation \: 3)$

{Now substituting the value of equation 3 in equation 1}

$\sf p(2p + \frac{py}{q} ) + qy = {p}^{2} - {q}^{2} \\ \\ \sf 2 {p}^{2} + \frac{ {p}^{2}y }{q} + qy = {p}^{2} - {q}^{2}$

{Taking 2p² on RHS and taking LCM on LHS}

$\sf \frac{ {p}^{2}y + {q}^{2} y }{q} = {p}^{2} - {q}^{2} - 2 {p}^{2} \\ \\ \sf \frac{y( {p}^{2} + {q}^{2} ) }{q} = - ( {p}^{2} + {q}^{2} )$

{Dividing (p² + q²) both sides}

$\sf \frac{y}{q} = - 1 \\ \\ \red{ \boxed {\bold{y = - q}} }$

{Now using value of y in equation 3}

$\sf x = 2p - \frac{pq}{q} \\ \\ \sf x = 2p - p \\ \\ \red{ \boxed{ \bold{ x = p}}}$