Question

Solve the system of linear equations
px + qy - p + q = 0
qx - py - p - q = 0

Answer 1

px + qy - p+ q =0 .......(1) ×q
qx - py - p - q = 0 .......(2) ×p

=> qpx + q^2y - qp + q^2= 0 .......(3)
qpx - p^2y - p^2 - pq= 0 ......... (4)

subtract (4) from (3)
=> q^2y + p^2y + p^2 + q^2 = 0

q^2y + p^2y = -p^2 -q^2
-y(-q^2-p^2) = -p^2 -q^2
-y= 1
y= -1

putting y= -1 in (1)

px + q (-1) - p+q =0
px - q -p + q = 0
px = p
X= 1

:. y= -1
x= 1


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Answer 2

Answer:

px+qy=p−q ............. (1)

qx−py=p+q .............(2)

Multiply equation (1) with p and equation (2) with q ,

p

2

x+pqy=p

2

−pq ........... (3)

q

2

x−pqy=pq+q

2

............ (4)

Add (3) and (4),

p

2

x+q

2

x=p

2

+q

2

x(p

2

+q

2

)=p

2

+q

2

⇒x=1

Put x=1 in equation (1) to get the value of y ,

p(1)+qy=p−q

p+qy=p−q

qy=−q

⇒y=−1

I hope this help you

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