Solve {l}{2a+3b+c=0}{a-b-c=4}{5a-3c+b=2}

$$\left. \begin{array} { l } { 2 a + 3 b + c = 0 } \\ { a - b - c = 4 } \\ { 5 a - 3 c + b = 2 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$b = -\frac{19}{4} = -4\frac{3}{4} = -4.75$$

$$c = \frac{21}{4} = 5\frac{1}{4} = 5.25$$

Short Steps Using Substitution

Solve 2a+3b+c=0 for c.

$$c=-2a-3b$$

Substitute -2a-3b for c in the second and third equation.

$$a-b-\left(-2a-3b\right)=4 5a-3\left(-2a-3b\right)+b=2$$

Solve these equations for b and a respectively.

$$b=-\frac{3}{2}a+2 a=\frac{2}{11}-\frac{10}{11}b$$

Substitute -\frac{3}{2}a+2 for b in the equation a=\frac{2}{11}-\frac{10}{11}b.

$$a=\frac{2}{11}-\frac{10}{11}\left(-\frac{3}{2}a+2\right)$$

Solve a=\frac{2}{11}-\frac{10}{11}\left(-\frac{3}{2}a+2\right) for a.

$$a=\frac{9}{2}$$

Substitute \frac{9}{2} for a in the equation b=-\frac{3}{2}a+2.

$$b=-\frac{3}{2}\times \frac{9}{2}+2$$

Calculate b from b=-\frac{3}{2}\times \frac{9}{2}+2.

$$b=-\frac{19}{4}$$

Substitute -\frac{19}{4} for b and \frac{9}{2} for a in the equation c=-2a-3b.

$$c=-2\times \frac{9}{2}-3\left(-\frac{19}{4}\right)$$

Calculate c from c=-2\times \frac{9}{2}-3\left(-\frac{19}{4}\right).

$$c=\frac{21}{4}$$

The system is now solved.

$$a=\frac{9}{2} b=-\frac{19}{4} c=\frac{21}{4}$$