Solve {l}{1+16+a+4b+c=0}{9+16+3a+4b+c=0}{16+a+4a+3b+c=0}

$$\left\{ \begin{array} { l } { 1 + 16 + a + 4 b + c = 0 } \\ { 9 + 16 + 3 a + 4 b + c = 0 } \\ { 16 + a + 4 a + 3 b + c = 0 } \end{array} \right.$$

Solve for a, b, c

$$a=-4$$

$$b=-17$$

$$c=55$$

Short Steps Using Substitution

Solve 1+16+a+4b+c=0 for a.

$$a=-17-4b-c$$

Substitute -17-4b-c for a in the second and third equation.

$$9+16+3\left(-17-4b-c\right)+4b+c=0 16-17-4b-c+4\left(-17-4b-c\right)+3b+c=0$$

Solve these equations for b and c respectively.

$$b=-\frac{13}{4}-\frac{1}{4}c c=-\frac{69}{4}-\frac{17}{4}b$$

Substitute -\frac{13}{4}-\frac{1}{4}c for b in the equation c=-\frac{69}{4}-\frac{17}{4}b.

$$c=-\frac{69}{4}-\frac{17}{4}\left(-\frac{13}{4}-\frac{1}{4}c\right)$$

Solve c=-\frac{69}{4}-\frac{17}{4}\left(-\frac{13}{4}-\frac{1}{4}c\right) for c.

$$c=55$$

Substitute 55 for c in the equation b=-\frac{13}{4}-\frac{1}{4}c.

$$b=-\frac{13}{4}-\frac{1}{4}\times 55$$

Calculate b from b=-\frac{13}{4}-\frac{1}{4}\times 55.

$$b=-17$$

Substitute -17 for b and 55 for c in the equation a=-17-4b-c.

$$a=-17-4\left(-17\right)-55$$

Calculate a from a=-17-4\left(-17\right)-55.

$$a=-4$$

The system is now solved.

$$a=-4 b=-17 c=55$$