Solve {l}{2=a+c}{3=a+b+2c}{0=-a-2b-2c}

$$\left. \begin{array} { l } { 2 = a + c } \\ { 3 = a + b + 2 c } \\ { 0 = - a - 2 b - 2 c } \end{array} \right.$$

Solve for a, c, b

$$a=-2$$

$$c=4$$

$$b=-3$$

Short Steps Using Substitution

Solve 2=a+c for a.

$$a=2-c$$

Substitute 2-c for a in the second and third equation.

$$3=2-c+b+2c 0=-\left(2-c\right)-2b-2c$$

Solve these equations for c and b respectively.

$$c=1-b b=-1-\frac{1}{2}c$$

Substitute 1-b for c in the equation b=-1-\frac{1}{2}c.

$$b=-1-\frac{1}{2}\left(1-b\right)$$

Solve b=-1-\frac{1}{2}\left(1-b\right) for b.

$$b=-3$$

Substitute -3 for b in the equation c=1-b.

$$c=1-\left(-3\right)$$

Calculate c from c=1-\left(-3\right).

$$c=4$$

Substitute 4 for c in the equation a=2-c.

$$a=2-4$$

Calculate a from a=2-4.

$$a=-2$$

The system is now solved.

$$a=-2 c=4 b=-3$$