Question

Simultaneous linear equations

1) Twice one number minus three times the second is equal to 2 ; and the sum of the number is 11 find number.

2). The sum of a two digit number and the number obtained on reversing the digits is 165 .If the digits differ by 3, find the number.

Answer 1

1)let numbers be x & y
3x-2y=2..................1
x+y=11...................2
..
Multiply equation 2 by 2
2x+2y=22
3x-2y+2x+2y=22+2
5x=24
x=24/5
plug the value of x in equation 1
you will get y =31/5

2)Let 1 digit be y.

:. Other digit will be y - 3

So, the no. is 10y + y - 3 = 11y-3

By reversing the digits the no. becomes 10(y - 3) + y = 11y - 30.

Sum of the 2 nos. = 165

11y - 3 + 11y - 30 = 165

22y - 33 = 165

22y = 198

y = 9

So, the no. is 96.

Since the digits differ by 3, they could be y and y +3.

So, the no. becomes 10y + y + 3 = 11y + 3

By reversing the digits, it becomes 10(y + 3) + y = 11y + 30

Therefore, 11y + 3 + 11y +30 = 165

22y +33 = 165

22y = 132

y = 6

Therefore the no. is 69