Solve {l}{9a+b-8c=0}{a+b-2c=0}{8a+2b-2c=0}

$$\left. \begin{array} { l } { 9 a + b - 8 c = 0 } \\ { a + b - 2 c = 0 } \\ { 8 a + 2 b - 2 c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=0$$

$$b=0$$

$$c=0$$

Short Steps Using Substitution

Solve 9a+b-8c=0 for b.

$$b=-9a+8c$$

Substitute -9a+8c for b in the second and third equation.

$$a-9a+8c-2c=0 8a+2\left(-9a+8c\right)-2c=0$$

Solve these equations for a and c respectively.

$$a=\frac{3}{4}c c=\frac{5}{7}a$$

Substitute \frac{3}{4}c for a in the equation c=\frac{5}{7}a.

$$c=\frac{5}{7}\times \frac{3}{4}c$$

Solve c=\frac{5}{7}\times \frac{3}{4}c for c.

$$c=0$$

Substitute 0 for c in the equation a=\frac{3}{4}c.

$$a=\frac{3}{4}\times 0$$

Calculate a from a=\frac{3}{4}\times 0.

$$a=0$$

Substitute 0 for a and 0 for c in the equation b=-9a+8c.

$$b=-9\times 0+8\times 0$$

Calculate b from b=-9\times 0+8\times 0.

$$b=0$$

The system is now solved.

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