Solve {l}{0=a-b+c}{0=1a+3b+c}{-3=c}

$$\left\{ \begin{array} { l } { 0 = a - b + c } \\ { 0 = 1 a + 3 b + c } \\ { - 3 = c } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b=0$$

$$c=-3$$

Short Steps Using Substitution

Reorder the equations.

$$-3=c 0=1a+3b+c 0=a-b+c$$

Solve -3=c for c.

$$c=-3$$

Substitute -3 for c in the second and third equation.

$$0=1a+3b-3 0=a-b-3$$

Solve these equations for b and a respectively.

$$b=-\frac{1}{3}a+1 a=b+3$$

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

$$a=-\frac{1}{3}a+1+3$$

Solve a=-\frac{1}{3}a+1+3 for a.

$$a=3$$

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

$$b=-\frac{1}{3}\times 3+1$$

Calculate b from b=-\frac{1}{3}\times 3+1.

$$b=0$$

The system is now solved.

$$a=3 b=0 c=-3$$