Solve {l}{a+2b+3c=1}{2a+2b+5c=0}{3a+5b+c=0}

$$\left\{ \begin{array} { l } { a + 2 b + 3 c = 1 } \\ { 2 a + 2 b + 5 c = 0 } \\ { 3 a + 5 b + c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a = -\frac{23}{15} = -1\frac{8}{15} \approx -1.533333333$$

$$b=\frac{13}{15}\approx 0.866666667$$

$$c=\frac{4}{15}\approx 0.266666667$$

Short Steps Using Substitution

Solve a+2b+3c=1 for a.

$$a=-2b-3c+1$$

Substitute -2b-3c+1 for a in the second and third equation.

$$2\left(-2b-3c+1\right)+2b+5c=0 3\left(-2b-3c+1\right)+5b+c=0$$

Solve these equations for b and c respectively.

$$b=-\frac{1}{2}c+1 c=-\frac{1}{8}b+\frac{3}{8}$$

Substitute -\frac{1}{2}c+1 for b in the equation c=-\frac{1}{8}b+\frac{3}{8}.

$$c=-\frac{1}{8}\left(-\frac{1}{2}c+1\right)+\frac{3}{8}$$

Solve c=-\frac{1}{8}\left(-\frac{1}{2}c+1\right)+\frac{3}{8} for c.

$$c=\frac{4}{15}$$

Substitute \frac{4}{15} for c in the equation b=-\frac{1}{2}c+1.

$$b=-\frac{1}{2}\times \frac{4}{15}+1$$

Calculate b from b=-\frac{1}{2}\times \frac{4}{15}+1.

$$b=\frac{13}{15}$$

Substitute \frac{13}{15} for b and \frac{4}{15} for c in the equation a=-2b-3c+1.

$$a=-2\times \frac{13}{15}-3\times \frac{4}{15}+1$$

Calculate a from a=-2\times \frac{13}{15}-3\times \frac{4}{15}+1.

$$a=-\frac{23}{15}$$

The system is now solved.

$$a=-\frac{23}{15} b=\frac{13}{15} c=\frac{4}{15}$$