Solve {l}{0=a-b+c}{6=a+6+c}{6=4a+2b+c}

$$\left\{ \begin{array} { l } { 0 = a - b + c } \\ { 6 = a + 6 + c } \\ { 6 = 4 a + 2 b + c } \end{array} \right.$$

Solve for a, b, c

$$a=2$$

$$b=0$$

$$c=-2$$

Short Steps Using Substitution

Solve 0=a-b+c for a.

$$a=b-c$$

Substitute b-c for a in the second and third equation.

$$6=b-c+6+c 6=4\left(b-c\right)+2b+c$$

Solve these equations for b and c respectively.

$$b=0 c=-2+2b$$

Substitute 0 for b in the equation c=-2+2b.

$$c=-2+2\times 0$$

Calculate c from c=-2+2\times 0.

$$c=-2$$

Substitute 0 for b and -2 for c in the equation a=b-c.

$$a=0-\left(-2\right)$$

Calculate a from a=0-\left(-2\right).

$$a=2$$

The system is now solved.

$$a=2 b=0 c=-2$$