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

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

Solve for b, a, c

$$b = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

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

$$c=\frac{1}{2}=0.5$$

Short Steps Using Substitution

Solve b+1=a for a.

$$a=b+1$$

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

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

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

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

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

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

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

$$b=-\frac{4}{3}$$

Substitute -\frac{4}{3} for b in the equation a=b+1.

$$a=-\frac{4}{3}+1$$

Calculate a from a=-\frac{4}{3}+1.

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

The system is now solved.

$$b=-\frac{4}{3} a=-\frac{1}{3} c=\frac{1}{2}$$