Answer to Question 7

Let a = #1, b= #2, c = #3
[1]:   
a + b + c = 2
[2]:   a + 5b + 5c = -10
[3]:    a + 2b + 3c = -3


Eliminate a:
subtract [1] from [2]:

4b + 4c = -12
Subtract [3] from [2]:
3b +2c = -7
Solve this new 2x2 system:
[4]:   4b + 4c = -12
[5]:    3b + 2c = -7
multiply [4] by 3 and [5] by 4, then subtract:
12b + 12c = -36
12b + 8c = 28
4c = -8
c = -2
Back-substitute into [4]:
4b + 4(-2) = -12
b = -1
Back-substitute into [1]:
a + (-1) + (-2) = 2
a = 5
Solution: (5, -1, -2)
CHECK:
[1]: 5 - 1 - 2 = 2
.2 = 2 check
[2]: 5 + 5(-1) + 5(-2) = -10
-10 = -10 check
[3]: 5 + 2(-1) + 3(-2) = -3
-3 = -3 check

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