Question

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

Answer 1

(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

(iii)

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