Problema Solution
The sum of three numbers is 8. The sum of twice the first number, 3 times the second number, and 4 times the third number is 23. The difference between 6 times the first number and second number is 23. Find the three numbers.
Answer provided by our tutors
let 'a' represent the first number, 'b' the second number and 'c' the third number, then we have these equations:
a+b+c=8
2a+3b+4c=23
6a-b=23
4*the first equation and subtract the second equation from it:
4a+4b+4c=32
2a+3b+4c=23
gives: 2a+b=9
two equations, two unknowns:
6a-b=23
2a+b=9
solving this gives:
a = 4
b = 1
solving for 'c' using equation 2:
2*4+3*1+4c=23
c = (23-8-3)/4 = 3
the numbers are 1, 3 and 4