Solve the following pair of linear equations:, 21x + 47y = 110, 47x + 21y = 162

Solution:

Given, the system of linear equations are

21x + 47y = 110 ------------------- (1)

47x + 21y = 162 ------------------- (2)

Adding (1) and (2),

21 x + 47y + 47x + 21y = 110 + 162

(21 + 47)x + (47 + 21)y = 272

68x + 68y = 272

Dividing by 68 on both sides,

x + y = 4 ----------- (3)

Subtracting (2) from (1),

(47x + 21y) - (21x + 47y) = 162 - 110

47x + 21y - 21x - 47y = 52

By grouping,

(47 - 21)x + (-47 + 21)y = 52

26x - 26y = 52

Dividing by 26 on both sides,

x - y = 2 ------------- (4)

Now, adding (3) and (4),

2x = 4 + 2

2x = 6

x = 6/2

x = 3

Put x = 3 in (1),

3 + y = 4

y = 4 - 3

y = 1

Therefore, the solution is x = 3 and y = 1

✦ Try This: Solve the following pair of linear equations: 11x + 37y = 110, 37x + 11y = 162

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 3

NCERT Exemplar Class 10 Maths Exercise 3.3 Sample Problem 2

Summary:

The solution to the system of equations 21x + 47y = 110; 47x + 21y = 162 are x = 3 and y = 1

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