Solve {l}{8a-3b+c=0}{4a+2b+c=a}{c=3}

$$\left\{ \begin{array} { l } { 8 a - 3 b + c = 0 } \\ { 4 a + 2 b + c = a } \\ { c = 3 } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{3}{5}=-0.6$$

$$b=-\frac{3}{5}=-0.6$$

$$c=3$$

Short Steps Using Substitution

Reorder the equations.

$$c=3 4a+2b+c=a 8a-3b+c=0$$

Substitute 3 for c in the second and third equation.

$$4a+2b+3=a 8a-3b+3=0$$

Solve these equations for b and a respectively.

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

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

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

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

$$a=-\frac{3}{5}$$

Substitute -\frac{3}{5} for a in the equation b=-\frac{3}{2}-\frac{3}{2}a.

$$b=-\frac{3}{2}-\frac{3}{2}\left(-\frac{3}{5}\right)$$

Calculate b from b=-\frac{3}{2}-\frac{3}{2}\left(-\frac{3}{5}\right).

$$b=-\frac{3}{5}$$

The system is now solved.

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