Solve {l}{a=b-c+2}{b=6-a-c}{c=3-4a-2b}

$$\left\{ \begin{array} { l } { a = b - c + 2 } \\ { b = 6 - a - c } \\ { c = 3 - 4 a - 2 b } \end{array} \right.$$

Solve for a, b, c

$$a = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

$$b=2$$

$$c = \frac{17}{3} = 5\frac{2}{3} \approx 5.666666667$$

Short Steps Using Substitution

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

$$b=6-\left(b-c+2\right)-c c=3-4\left(b-c+2\right)-2b$$

Solve these equations for b and c respectively.

$$b=2 c=\frac{5}{3}+2b$$

Substitute 2 for b in the equation c=\frac{5}{3}+2b.

$$c=\frac{5}{3}+2\times 2$$

Calculate c from c=\frac{5}{3}+2\times 2.

$$c=\frac{17}{3}$$

Substitute 2 for b and \frac{17}{3} for c in the equation a=b-c+2.

$$a=2-\frac{17}{3}+2$$

Calculate a from a=2-\frac{17}{3}+2.

$$a=-\frac{5}{3}$$

The system is now solved.

$$a=-\frac{5}{3} b=2 c=\frac{17}{3}$$