Solve {l}{1+a+b+c=-2}{3+2a+b=0}{2-4/3a+b=0}

$$\left. \begin{array} { l } { 1 + a + b + c = - 2 } \\ { 3 + 2 a + b = 0 } \\ { 2 - \frac { 4 } { 3 } a + b = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{3}{10}=-0.3$$

$$b = -\frac{12}{5} = -2\frac{2}{5} = -2.4$$

$$c=-\frac{3}{10}=-0.3$$

Short Steps Using Substitution

Solve 1+a+b+c=-2 for a.

$$a=-3-b-c$$

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

$$3+2\left(-3-b-c\right)+b=0 2-\frac{4}{3}\left(-3-b-c\right)+b=0$$

Solve these equations for b and c respectively.

$$b=-3-2c c=-\frac{9}{2}-\frac{7}{4}b$$

Substitute -3-2c for b in the equation c=-\frac{9}{2}-\frac{7}{4}b.

$$c=-\frac{9}{2}-\frac{7}{4}\left(-3-2c\right)$$

Solve c=-\frac{9}{2}-\frac{7}{4}\left(-3-2c\right) for c.

$$c=-\frac{3}{10}$$

Substitute -\frac{3}{10} for c in the equation b=-3-2c.

$$b=-3-2\left(-\frac{3}{10}\right)$$

Calculate b from b=-3-2\left(-\frac{3}{10}\right).

$$b=-\frac{12}{5}$$

Substitute -\frac{12}{5} for b and -\frac{3}{10} for c in the equation a=-3-b-c.

$$a=-3-\left(-\frac{12}{5}\right)-\left(-\frac{3}{10}\right)$$

Calculate a from a=-3-\left(-\frac{12}{5}\right)-\left(-\frac{3}{10}\right).

$$a=-\frac{3}{10}$$

The system is now solved.

$$a=-\frac{3}{10} b=-\frac{12}{5} c=-\frac{3}{10}$$