Question

.. px + qy =p-q
qx - Py =p+q. with eliminated method​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:px + qy = p - q - - - (1)$

and

$\rm :\longmapsto\:qx - py = p + q - - - (2)$

Now, to solve these equations with Method of Eliminations, we have to first make the leading coefficients of x, same in both the equations.

So,

Multiply equation (1) by q, we get

$\rm :\longmapsto\:pqx + {q}^{2} y = pq - {q}^{2} - - - (3)$

Now,

Multiply equation (2) by p, we get

$\rm :\longmapsto\:qpx - {p}^{2} y = {p}^{2} + qp - - - (3)$

Now, Subtract equation (2) from equation (1), we get

$\rm :\longmapsto\: {q}^{2}y + {p}^{2}y = - {q}^{2} - {p}^{2}$

$\rm :\longmapsto\: ({q}^{2} + {p}^{2})y = - ({q}^{2} + {p}^{2})$

$\bf\implies \:y = - 1$

On substituting y = - 1, in equation (1), we get

$\rm :\longmapsto\:px + q( - 1)= p - q$

$\rm :\longmapsto\:px - q= p - q$

$\rm :\longmapsto\:px= p$

$\bf :\longmapsto\:x= 1$

Verification :-

Consider equation (1),

$\rm :\longmapsto\:px + qy= p - q$

On Substituting the values of x and y, we have

$\rm :\longmapsto\:p(1) + q( - 1)= p - q$

$\rm :\longmapsto\:p - q= p - q$

Verified

Consider equation (2),

$\rm :\longmapsto\:qx - py = p + q$

On substituting the values of x and y, we have

$\rm :\longmapsto\:q(1) - p( - 1) = p + q$

$\rm :\longmapsto\:q + p = p + q$

Verified