Solve {l}{b+1=a}{c+b+1/2=0}{6c+3b+1=0}

$$\left. \begin{array} { l } { b + 1 = a } \\ { c + b + \frac { 1 } { 2 } = 0 } \\ { 6 c + 3 b + 1 = 0 } \end{array} \right.$$

Solve for b, a, c

$$b=-\frac{2}{3}\approx -0.666666667$$

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

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

Short Steps Using Substitution

Solve b+1=a for a.

$$a=b+1$$

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

$$b=-c-\frac{1}{2} c=-\frac{1}{2}b-\frac{1}{6}$$

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

$$c=-\frac{1}{2}\left(-c-\frac{1}{2}\right)-\frac{1}{6}$$

Solve c=-\frac{1}{2}\left(-c-\frac{1}{2}\right)-\frac{1}{6} for c.

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

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

$$b=-\frac{1}{6}-\frac{1}{2}$$

Calculate b from b=-\frac{1}{6}-\frac{1}{2}.

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

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

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

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

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

The system is now solved.

$$b=-\frac{2}{3} a=\frac{1}{3} c=\frac{1}{6}$$