Problema Solution
The sum of three numbers is 16. The sum of twice the first number, 3 times the second number, and 4 times the third number is 49. The difference between 5 times the first number and the second number is 39. Find the three numbers.
Answer provided by our tutors
let the numbers be represented by the variables a, b, and c, then we have these equations:
a+b+c=16
2a+3b+4c=49
5a-b=39
we can eliminate 'c' from the first equation by adding negative four times the first equation to the second equation:
-4a-4b-4c=-64
2a+3b+4c=49
giving:
-2a-b=-15
solving these equations:
-2a-b=-15
5a-b=39
gives fractional answers for 'a' and 'b:
a=54/7
b=-3/7
therefore, this problem no doubt contains logical errors since these types of problems should result in integer solutions