Question

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Solve the given equations.

=> px + qy = p - q

=> qx - py = p + q

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Answer 1

Answer:

Step-by-step explanation:

This can be done by the sole and the easiest method of reduction - elimination method.

https://brainly.in/question/37978634 - A similar question asked with almost similar procedures.

$\sf{px + qy = p - q}$ -------(1)

$\sf{qx - py = p + q}$ --------(2)

What we can do is to eliminate the variable, we can multiply (1) with p and (2) with q

$\sf{p(px + qy = p - q)}$

$\sf{p^2 x + qpy = p(p - q)}$ ----(3)

$\sf{q(qx - py = p + q)}$

$\sf{q^2 x - pqy = q(p + q)}$ -----(4)

Taking equations (3) and (4), we can add them.

$\sf{qpy + p^2 x - pqy + q^2 x = q(p + q) + p(p - q)}$

qpy and pqy cancel with opposite signs get cancelled, we get:-

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

Opening brackets, we get

$\sf{ q^2 x + p^2 x = qp + q^2 + p^2 - qp}$

Again, qp and pq get cancelled due to opposite signs.

$\sf{ x(q^2 + p^2) = (q^2 + p^2)}$

q² + p² gets cancelled,

x= 1 is the 1st answer.

Now, as we need y, substituting value of x in equation (1)

$\sf{p + qy = p - q}$

$\sf{qy = - q}$

y = -1 is the second answer.

So, x = 1 and y = -1.

Answer 2

$\underbrace \text{Answer:}$

Given:

px+qy=p-q ..(1)

qx-py=p+q ..(2)

➼ Multiply equation (1) with p and equation (2) with q.

$\tt{p(px + qy = p - q)}$

$\implies \tt{ {p}^{2}x + pqy = p ^{2} + pq }..(3)$

$\tt{q(qx - py = p + q)}$

$\implies \tt{ {q}^{2} x - qpy = qp + {q}^{2} }..(4)$

➼ Add (3) and (4),

$\tt{ {p}^{2}x + {q}^{2}x = {p}^{2} + {q}^{2} }$

$\implies \tt{x( {p}^{2} + {q}^{2} ) = {p}^{2} + {q}^{2} }$

$\implies \bf{ \red{x = 1}}$

➼ Now, put the value of x in equation (1), to get the value of y.

$\tt{p(1) + qy = p - q}$

$\implies \tt{p + qy = p - q}$

$\implies \tt{qy = \cancel{p }- \cancel p - q}$

$\implies \tt{qy = - q}$

$\implies \bf{ \red{y = - 1}}$

$⛬ \: \text{The value of x is \red{1} and y is \pink{ - 1}}.$