Solve {r}{a+b+2c=2}{2b+2c=1}{-a+b+3c=0}

$$\left. \begin{array} { r } { a + b + 2 c = 2 } \\ { 2 b + 2 c = 1 } \\ { - a + b + 3 c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

$$b=\frac{1}{6}\approx 0.166666667$$

$$c=\frac{1}{3}\approx 0.333333333$$

Short Steps Using Substitution

Solve a+b+2c=2 for a.

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

Substitute -b-2c+2 for a in the equation -a+b+3c=0.

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

Solve the second equation for b and the third equation for c.

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

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

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

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

$$c=\frac{1}{3}$$

Substitute \frac{1}{3} for c in the equation b=-c+\frac{1}{2}.

$$b=-\frac{1}{3}+\frac{1}{2}$$

Calculate b from b=-\frac{1}{3}+\frac{1}{2}.

$$b=\frac{1}{6}$$

Substitute \frac{1}{6} for b and \frac{1}{3} for c in the equation a=-b-2c+2.

$$a=-\frac{1}{6}-2\times \frac{1}{3}+2$$

Calculate a from a=-\frac{1}{6}-2\times \frac{1}{3}+2.

$$a=\frac{7}{6}$$

The system is now solved.

$$a=\frac{7}{6} b=\frac{1}{6} c=\frac{1}{3}$$