Question

Find two numbers such that the sum of twice the first and thrice the second is 103 and 4 times the first exceed 7 times the second by 11

Answer 1

29and 15 are ur no. s

Answer 2

let the nos be x and y.
A/q 2x + 3y = 103
=> 2x = 103-3y
A/q 4x=7y + 11
=> 2(103-3y)=7y+11
=> 206 -6y =7y +11
=> 206-11=7y+6y
=> 13y =195
=> y =15
=> x = (103-3*15)/2
=103-45)/2
= 58/2
=29

thus, x=29 and y=15