Solve {l}{0=16a-4b+c}{0=4a+2b+c}{-4=c}

$$\left\{ \begin{array} { l } { 0 = 16 a - 4 b + c } \\ { 0 = 4 a + 2 b + c } \\ { - 4 = c } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{1}{2}=0.5$$

$$b=1$$

$$c=-4$$

Short Steps Using Substitution

Reorder the equations.

$$-4=c 0=4a+2b+c 0=16a-4b+c$$

Solve -4=c for c.

$$c=-4$$

Substitute -4 for c in the second and third equation.

$$0=4a+2b-4 0=16a-4b-4$$

Solve these equations for b and a respectively.

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

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

$$a=\frac{1}{4}+\frac{1}{4}\left(2-2a\right)$$

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

$$a=\frac{1}{2}$$

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

$$b=2-2\times \frac{1}{2}$$

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

$$b=1$$

The system is now solved.

$$a=\frac{1}{2} b=1 c=-4$$