Solve {l}{0=16a+4b+c}{0=4a-2b+c}{-9/2=a+b+c}

$$\left\{ \begin{array} { l } { 0 = 16 a + 4 b + c } \\ { 0 = 4 a - 2 b + c } \\ { - \frac { 9 } { 2 } = a + b + c } \end{array} \right.$$

Solve for a, b, c

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

$$b=-1$$

$$c=-4$$

Short Steps Using Substitution

Solve 0=16a+4b+c for c.

$$c=-16a-4b$$

Substitute -16a-4b for c in the second and third equation.

$$0=4a-2b-16a-4b -\frac{9}{2}=a+b-16a-4b$$

Solve these equations for b and a respectively.

$$b=-2a a=\frac{3}{10}-\frac{1}{5}b$$

Substitute -2a for b in the equation a=\frac{3}{10}-\frac{1}{5}b.

$$a=\frac{3}{10}-\frac{1}{5}\left(-2\right)a$$

Solve a=\frac{3}{10}-\frac{1}{5}\left(-2\right)a for a.

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

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

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

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

$$b=-1$$

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

$$c=-16\times \frac{1}{2}-4\left(-1\right)$$

Calculate c from c=-16\times \frac{1}{2}-4\left(-1\right).

$$c=-4$$

The system is now solved.

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