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

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

Solve for a, b, c

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

$$b=1$$

$$c=4$$

Short Steps Using Substitution

Reorder the equations.

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

Substitute 4 for c in the second and third equation.

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

Solve these equations for b and a respectively.

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

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

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

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

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

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

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

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

$$b=1$$

The system is now solved.

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