Solve {c}{a+2b+3c=0}{2a+5b+7c=0}{3a+7b+(10+8)c=0}

$$\left\{ \begin{array} { c } { a + 2 b + 3 c = 0 } \\ { 2 a + 5 b + 7 c = 0 } \\ { 3 a + 7 b + ( 10 + 8 ) c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=0$$

$$b=0$$

$$c=0$$

Short Steps Using Substitution

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

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

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

$$2\left(-2b-3c\right)+5b+7c=0 3\left(-2b-3c\right)+7b+\left(10+8\right)c=0$$

Solve these equations for b and c respectively.

$$b=-c c=-\frac{1}{9}b$$

Substitute -c for b in the equation c=-\frac{1}{9}b.

$$c=-\frac{1}{9}\left(-1\right)c$$

Solve c=-\frac{1}{9}\left(-1\right)c for c.

$$c=0$$

Substitute 0 for c in the equation b=-c.

$$b=-0$$

Calculate b from b=-0.

$$b=0$$

Substitute 0 for b and 0 for c in the equation a=-2b-3c.

$$a=-2\times 0-3\times 0$$

Calculate a from a=-2\times 0-3\times 0.

$$a=0$$

The system is now solved.

$$a=0 b=0 c=0$$