Solve {l}{a-b+c=0}{16a+4b+c=0}{4a+2b+c=-6}

$$\left\{ \begin{array} { l } { a - b + c = 0 } \\ { 16 a + 4 b + c = 0 } \\ { 4 a + 2 b + c = - 6 } \end{array} \right.$$

Solve for a, b, c

$$a=1$$

$$b=-3$$

$$c=-4$$

Short Steps Using Substitution

Solve a-b+c=0 for a.

$$a=b-c$$

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

$$16\left(b-c\right)+4b+c=0 4\left(b-c\right)+2b+c=-6$$

Solve these equations for b and c respectively.

$$b=\frac{3}{4}c c=2+2b$$

Substitute \frac{3}{4}c for b in the equation c=2+2b.

$$c=2+2\times \frac{3}{4}c$$

Solve c=2+2\times \frac{3}{4}c for c.

$$c=-4$$

Substitute -4 for c in the equation b=\frac{3}{4}c.

$$b=\frac{3}{4}\left(-4\right)$$

Calculate b from b=\frac{3}{4}\left(-4\right).

$$b=-3$$

Substitute -3 for b and -4 for c in the equation a=b-c.

$$a=-3-\left(-4\right)$$

Calculate a from a=-3-\left(-4\right).

$$a=1$$

The system is now solved.

$$a=1 b=-3 c=-4$$