Solve the following pair of linear equations:

(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

x == 1

From equation (1),

we obtain p (1) + qy = p - q
qy = - q,so, y = - 1

(ii) ax + by = c … (1) bx + ay = 1 + c … (2)
Multiplying equation (1) by a and equation (2) by b, we obtain:

Subtracting equation (4) from equation (3),

From equation (1), we obtain ax + by = c

(iii)

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 - b²a²- b² + (a + b)y = a² -2ab - b²

(a + b)y = - 2ab

y = -2ab/(a + b)

(v)152x – 378y = –74… (1)
–378x + 152y = –604 … (2)
Adding the equations (1) and (2), we obtain:
–226x – 226y = –678

Subtracting the equation (2) from equation (1), we obtain:
530x – 530y = 530

Adding equations (3) and (4), we obtain:
2x = 4

x = 2
Substituting the value of x in equation (3), we obtain:
y = 1