solve the following pair of linear equations

21x + 47y = 110

47 x + 21 y = 162

We have
21x + 47y = 110 (1)
47x + 21y = 162 (2)
Multiplying Equation (1) by 47 and Equation (2) by 21, we get
987x + 2209 y = 5170 (3)
987x + 441y = 3402 (4)
Subtracting Equation (4) from Equation (3), we get
1768y = 1768
or y = 1
Substituting the value of y in Equation (1), we get
21x + 47 = 110
or 21x = 63
or x = 3
So, x = 3, y = 1