Solve {l}{a+2b+3c-3=0}{3a+b+2c-7=0}{2a+3b+c-2=0}

$$\left\{ \begin{array} { l } { a + 2 b + 3 c - 3 = 0 } \\ { 3 a + b + 2 c - 7 = 0 } \\ { 2 a + 3 b + c - 2 = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=2$$

$$b=-1$$

$$c=1$$

Short Steps Using Substitution

Solve a+2b+3c-3=0 for a.

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

Substitute -2b-3c+3 for a in the second and third equation.

$$3\left(-2b-3c+3\right)+b+2c-7=0 2\left(-2b-3c+3\right)+3b+c-2=0$$

Solve these equations for b and c respectively.

$$b=-\frac{7}{5}c+\frac{2}{5} c=-\frac{1}{5}b+\frac{4}{5}$$

Substitute -\frac{7}{5}c+\frac{2}{5} for b in the equation c=-\frac{1}{5}b+\frac{4}{5}.

$$c=-\frac{1}{5}\left(-\frac{7}{5}c+\frac{2}{5}\right)+\frac{4}{5}$$

Solve c=-\frac{1}{5}\left(-\frac{7}{5}c+\frac{2}{5}\right)+\frac{4}{5} for c.

$$c=1$$

Substitute 1 for c in the equation b=-\frac{7}{5}c+\frac{2}{5}.

$$b=-\frac{7}{5}+\frac{2}{5}$$

Calculate b from b=-\frac{7}{5}+\frac{2}{5}.

$$b=-1$$

Substitute -1 for b and 1 for c in the equation a=-2b-3c+3.

$$a=-2\left(-1\right)-3+3$$

Calculate a from a=-2\left(-1\right)-3+3.

$$a=2$$

The system is now solved.

$$a=2 b=-1 c=1$$