Solve {l}{a+2b+3c=2}{3a+b+15c=18}{4a-9c=17}

$$\left. \begin{array} { l } { a + 2 b + 3 c = 2 } \\ { 3 a + b + 15 c = 18 } \\ { 4 a - 9 c = 17 } \end{array} \right.$$

Solve for a, b, c

$$a=5$$

$$b=-2$$

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

Short Steps Using Substitution

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

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

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

$$3\left(-2b-3c+2\right)+b+15c=18 4\left(-2b-3c+2\right)-9c=17$$

Solve these equations for b and c respectively.

$$b=-\frac{12}{5}+\frac{6}{5}c c=-\frac{8}{21}b-\frac{3}{7}$$

Substitute -\frac{12}{5}+\frac{6}{5}c for b in the equation c=-\frac{8}{21}b-\frac{3}{7}.

$$c=-\frac{8}{21}\left(-\frac{12}{5}+\frac{6}{5}c\right)-\frac{3}{7}$$

Solve c=-\frac{8}{21}\left(-\frac{12}{5}+\frac{6}{5}c\right)-\frac{3}{7} for c.

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

Substitute \frac{1}{3} for c in the equation b=-\frac{12}{5}+\frac{6}{5}c.

$$b=-\frac{12}{5}+\frac{6}{5}\times \frac{1}{3}$$

Calculate b from b=-\frac{12}{5}+\frac{6}{5}\times \frac{1}{3}.

$$b=-2$$

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

$$a=-2\left(-2\right)-3\times \frac{1}{3}+2$$

Calculate a from a=-2\left(-2\right)-3\times \frac{1}{3}+2.

$$a=5$$

The system is now solved.

$$a=5 b=-2 c=\frac{1}{3}$$