Solve {c}{36a+6b+c=0}{c=3}{16a-4b+c=0}

$$\left\{ \begin{array} { c } { 36 a + 6 b + c = 0 } \\ { c = 3 } \\ { 16 a - 4 b + c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=-\frac{1}{8}=-0.125$$

$$b=\frac{1}{4}=0.25$$

$$c=3$$

Short Steps Using Substitution

Reorder the equations.

$$c=3 36a+6b+c=0 16a-4b+c=0$$

Substitute 3 for c in the second and third equation.

$$36a+6b+3=0 16a-4b+3=0$$

Solve these equations for b and a respectively.

$$b=-\frac{1}{2}-6a a=\frac{1}{4}b-\frac{3}{16}$$

Substitute -\frac{1}{2}-6a for b in the equation a=\frac{1}{4}b-\frac{3}{16}.

$$a=\frac{1}{4}\left(-\frac{1}{2}-6a\right)-\frac{3}{16}$$

Solve a=\frac{1}{4}\left(-\frac{1}{2}-6a\right)-\frac{3}{16} for a.

$$a=-\frac{1}{8}$$

Substitute -\frac{1}{8} for a in the equation b=-\frac{1}{2}-6a.

$$b=-\frac{1}{2}-6\left(-\frac{1}{8}\right)$$

Calculate b from b=-\frac{1}{2}-6\left(-\frac{1}{8}\right).

$$b=\frac{1}{4}$$

The system is now solved.

$$a=-\frac{1}{8} b=\frac{1}{4} c=3$$