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

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

Solve for a, b, c

$$a=1$$

$$b=-2$$

$$c=-8$$

Short Steps Using Substitution

Solve 4a-2b+c=0 for c.

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

Substitute -4a+2b for c in the second and third equation.

$$a+b-4a+2b=-9 16a+4b-4a+2b=0$$

Solve these equations for b and a respectively.

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

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

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

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

$$a=1$$

Substitute 1 for a in the equation b=-3+a.

$$b=-3+1$$

Calculate b from b=-3+1.

$$b=-2$$

Substitute -2 for b and 1 for a in the equation c=-4a+2b.

$$c=-4+2\left(-2\right)$$

Calculate c from c=-4+2\left(-2\right).

$$c=-8$$

The system is now solved.

$$a=1 b=-2 c=-8$$