# 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) x/a - y/b = 0; 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

**Solution:**

(i)

px + qy = p - q ....(1)

qx - py = p + q ....(2)

Multiplying equation (1) by p and equation (2) by q, we obtain

p² x + pqy = p² - pq ....(3)

q² x - pqy = pq + q² ....( 4)

Adding equations (3) and (4), we obtain

p²x + q²x = p² + q²

( p² + q²) x = p² + q²

x = (p² + q²)/p²+ q²

Substituting

x = 1

x =1 in equation (1), we obtain

p × 1 + qy = p - q

qy = - q

y = - 1

Therefore, x = 1 and y = - 1

(ii)

ax + by = c ....(1)

bx + ay = 1+ c ....(2)

Multiplying equation (1) by a and equation (2) by b , we obtain

a² x + aby = ac ....(3)

b^{2} x + aby = b + bc ....(4)

Subtracting equation (4) from equation (3),

(a² - b²) x = ac - bc - b

x = c(a - b) - b/(a² - b²)

Substituting x = c(a - b) - b/(a² - b²) in equation (1), we obtain

ax + by = c

x[ c(a - b) - b/(a² - b²) + by] = c

ac(a - b) - ab/(a² - b²) + by = c

by = c - ac(a - b) - ab/(a² - b²)

by = a²c - b²c - a²c + abc + ab/(a² - b²)

by = abc - b²c + ab/(a² - b²)

by = bc(a - b) + ab/(a² --b²)

by = b [c(a - b) + a]/(a² - b²)

y = c(a - b) + a/(a² - b²)

Therefore, x = c(a - b) - b/(a² - b²) and y = c(a - b) + a/(a² - b²)

(iii)

x/a - y/b = 0 ....(1)

ax + by = a² + b² ....(2)

By solving equation (1), we obtain

x/a - y/b = 0

x = ay/b ....(4)

Substituting x = ay/b in equation (2), we obtain

a (ay/b) + by = a² + b²

(a² y + b² y)/b = a² + b²

(a² + b²) y = b (a² + b²)

y = b

Substituting y = b in equation (3), we obtain

x = (a × b)/b

x = a

Therefore, x= a and y = b

(iv)

(a - b)x + (a + b) y = a² - 2ab - b² ....(1)

(a + b)(x + y) = a² + b² ....(2)

By solving equation (2), we obtain

(a + b)(x + y) = a² + b²

(a + b)x + (a + b) y = a² + b² ....(3)

Subtracting equation (3) from (1), we obtain

(a - b)x - (a + b)x = (a² - 2ab - b²) - (a² + b²)

[(a - b) - (a + b)] x = a² - 2ab - b² - a² - b²

[a - b - a - b] x = -2ab - 2b²

-2bx = -2b (a + b)

x = (a + b)

Substituting x = (a + b) in 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 = a² - 2ab - b² - (a² - b²)

(a + b) y = a² - 2ab - b² - a² + b²

y = -2ab/(a + b)

(v)

152x - 378y = -74 ....(1)

- 378x + 152y = - 604 ....(2)

Adding equations (1) and (2), we obtain

-226x - 226y = -678

-226( x + y ) = -678

x + y = 3 ...(3)

Subtracting equation (2) from (1), we obtain

530x - 530y = 530

530 ( x - y ) = 530

x - y = 1 ....(4)

Adding equations (3) and (4), we obtain

2x = 4

x = 2

Substituting x = 2 in equation (3), we obtain

2 + y = 3

y = 1

Therefore, x = 2 and y = 1

**Video Solution:**

## 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) x/a - y/b = 0; 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

### Class 10 Maths NCERT Solutions - Chapter 3 Exercise 3.7 Question 7:

For the following linear equations the solutions are as follows, for the first equation the solution is (1, -1) for the second equation the solution is (x = c(a - b) - b , y = c(a - b) + a) for the third equation the solution is ( x = a, y = b) and for the fourth equation the solution is ( x = (a + b), y = -2ab/(a + b) ) and for the fifth equation the solution is ( x = 2, y = 1)