Solve the following pair of linear equations.
(i) px + qy = p − q
qx − py = p + q
(ii) ax + by = c
bx + ay = 1 + c
(iii) 
ax + by = a2 + b2
(iv) (a − b) x + (a + b) y = a2− 2ab − b2
(a + b) (x + y) = a2 + b2
(v) 152x − 378y = − 74
− 378x + 152y = − 604
(i)px + qy = p − q … (1)
qx − py = p + q … (2)
Multiplying equation (1) by p and equation (2) by q, we obtain
p2x + pqy = p2 − pq … (3)
q2x − pqy = pq + q2 … (4)
Adding equations (3) and (4), we obtain
p2x + q2 x = p2 + q2
(p2 + q2) x = p2 + q2
From equation (1), we obtain
p (1) + qy = p − q
qy = − q
y = − 1
(ii)ax + by = c … (1)
bx + ay = 1 + c … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain
a2x + aby = ac … (3)
b2x + aby = b + bc … (4)
Subtracting equation (4) from equation (3),
(a2 − b2) x = ac − bc − b
From equation (1), we obtain
ax + by = c
Or, bx − ay = 0 … (1)
ax + by = a2 + b2 … (2)
Multiplying equation (1) and (2) by b and a respectively, we obtain
b2x − aby = 0 … (3)
a2x + aby = a3 + ab2 … (4)
Adding equations (3) and (4), we obtain
b2x + a2x = a3 + ab2
x (b2 + a2) = a (a2 + b2)
x = a
By using (1), we obtain
b (a) − ay = 0
ab − ay = 0
ay = ab
y = b
(iv) (a − b) x + (a + b) y = a2− 2ab − b2 … (1)
(a + b) (x + y) = a2 + b2
(a + b) x + (a + b) y = a2 + b2 … (2)
Subtracting equation (2) from (1), we obtain
(a − b) x − (a + b) x = (a2 − 2ab − b2) − (a2 + b2)
(a − b − a − b) x = − 2ab − 2b2
− 2bx = − 2b (a + b)
x = a + b
Using equation (1), we obtain
(a − b) (a + b) + (a + b) y = a2 − 2ab − b2
a2 − b2 + (a + b) y = a2− 2ab − b2
(a + b) y = − 2ab
(v) 152x − 378y = − 74
76x − 189y = − 37
… (1)
− 378x + 152y = − 604
− 189x + 76y = − 302 … (2)
Substituting the value of x in equation (2), we obtain
− (189)2 y + 189 × 37 + (76)2 y = − 302 × 76
189 × 37 + 302 × 76 = (189)2 y − (76)2 y
6993 + 22952 = (189 − 76) (189 + 76) y
29945 = (113) (265) y
y = 1
From equation (1), we obtain
