Question 251605
I need to solve this problem with a system of three equations for age: The sum of the ages of three brothers, Andre, Brad, and Carlos is 48. In ten years, twice Brad's age will be 3 years less than the sum of his brothers ages. Six years ago, Carlos' age was four years younger than the sum of his brothers ages. How old are the brothers now?

i failed with these equations:
a+b+c=48
b+10=(a+c)-7
c-6=(b+a)-6


Let Andre's, Brad's, and Carlos' ages be A, B, & C, respectively


Then we get: A + B + C  =  48


Since in ten years, twice Brad's age will be 3 years less than the sum of his brothers ages, then 2(B + 10) = (A + 10) + (C + 10) - 3, or, - A + 2B - C = - 3


Since six years ago, Carlos' age was four years younger than the sum of his brothers ages, then C - 6 = (A - 6) + (B - 6) - 4, or, - A - B + C = - 10


We now have the following system of equations:

----------   A +  B + C  =  48 ------ eq (i)

--------   - A + 2B - C = - 3 ------- eq (ii)

-------- - A - B + C = - 10 ------- eq (iii)


Add eqs (i) and (ii) to get: 3B  =  45


Therefore B = 15


Add eqs (i) and (iii) to get: 2C  =  38


Therefore, C = 19


Substitute 15 for B, and 19 for C in eq (i) to get: A + 15 + 19 = 48


Therefore, A  =  14


Therefore, Andre, Brad, and Carlos are {{{highlight_green(14_15_and_19)}}} years-old, respectively.

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You can do the check yourself. I'm sure you can at least handle the check.