\left\{ \begin{array} { l } { 5 = a + c } \\ { 10 = 2 a + 3 b + c } \\ { 11 = 3 a + 2 b + c } \end{array} \right.
Solve for a, c, b
a=2
c=3
b=1
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a=5-c
Solve 5=a+c for a.
10=2\left(5-c\right)+3b+c 11=3\left(5-c\right)+2b+c
Substitute 5-c for a in the second and third equation.
c=3b b=-2+c
Solve these equations for c and b respectively.
b=-2+3b
Substitute 3b for c in the equation b=-2+c.
b=1
Solve b=-2+3b for b.
c=3\times 1
Substitute 1 for b in the equation c=3b.
c=3
Calculate c from c=3\times 1.
a=5-3
Substitute 3 for c in the equation a=5-c.
a=2
Calculate a from a=5-3.
a=2 c=3 b=1
The system is now solved.
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