SOLUTION: 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

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
Found 3 solutions by stanbon, richwmiller, MathTherapy:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
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.
a + b + c = 48
---
In ten years, twice Brad's age will be 3 years less than the sum of his brothers ages. six years ago,
2(b+10) = (a-6) + (c-6)-3
2b+20 = a+c-15
a - 2b +c = 35
---
Carlos's age was four years younger than the sum of his brothers ages.
c = a + b - 4
a + b - c = 4
------------------------
how old are the brothers now?
---------
Gather the equations:
a + b + c = 48
a - 2b + c = 35
a + b - c = 4
----------------------------
I used a matrix method and got:
a = 21 2/3
b = 4 1/3
c = 22
=============
Cheers,
Stan H.

Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
Sorry Stanborn,
a+b+c=48
2(b+10)=(a+10+c+10)-3
(c-6)=(a-6)+(b-6)+4

a = 10, b = 15, c = 23
check
10+15+23=48
2*(15+10)=10+10+23+10-3
23-6=(10-6)+(15-6)+4
all ok

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
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 years-old, respectively.
---------
Check
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You can do the check yourself. I'm sure you can at least handle the check.
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