Solve {l}{2=c}{18=16a+4b+c}{8=-a-b+c}

$$\left. \begin{array}{l}{ 2 = c }\\{ 18 = 16 a + 4 b + c }\\{ 8 = - a - b + c }\end{array} \right.$$

Solve for c, a, b

$$c=2$$

$$a = \frac{10}{3} = 3\frac{1}{3} \approx 3.333333333$$

$$b = -\frac{28}{3} = -9\frac{1}{3} \approx -9.333333333$$

Short Steps Using Substitution

Solve 2=c for c.

$$c=2$$

Substitute 2 for c in the second and third equation.

$$18=16a+4b+2 8=-a-b+2$$

Solve these equations for a and b respectively.

$$a=1-\frac{1}{4}b b=-a-6$$

Substitute 1-\frac{1}{4}b for a in the equation b=-a-6.

$$b=-\left(1-\frac{1}{4}b\right)-6$$

Solve b=-\left(1-\frac{1}{4}b\right)-6 for b.

$$b=-\frac{28}{3}$$

Substitute -\frac{28}{3} for b in the equation a=1-\frac{1}{4}b.

$$a=1-\frac{1}{4}\left(-\frac{28}{3}\right)$$

Calculate a from a=1-\frac{1}{4}\left(-\frac{28}{3}\right).

$$a=\frac{10}{3}$$

The system is now solved.

$$c=2 a=\frac{10}{3} b=-\frac{28}{3}$$