Solve {l}{4=4a+2b+c}{32=16a+4b+c}{108=36a+6b+c}

$$\left\{ \begin{array} { l } { 4 = 4 a + 2 b + c } \\ { 32 = 16 a + 4 b + c } \\ { 108 = 36 a + 6 b + c } \end{array} \right.$$

Solve for a, b, c

$$a=6$$

$$b=-22$$

$$c=24$$

Short Steps Using Substitution

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

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

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

$$32=16a+4b+4-4a-2b 108=36a+6b+4-4a-2b$$

Solve these equations for b and a respectively.

$$b=14-6a a=\frac{13}{4}-\frac{1}{8}b$$

Substitute 14-6a for b in the equation a=\frac{13}{4}-\frac{1}{8}b.

$$a=\frac{13}{4}-\frac{1}{8}\left(14-6a\right)$$

Solve a=\frac{13}{4}-\frac{1}{8}\left(14-6a\right) for a.

$$a=6$$

Substitute 6 for a in the equation b=14-6a.

$$b=14-6\times 6$$

Calculate b from b=14-6\times 6.

$$b=-22$$

Substitute -22 for b and 6 for a in the equation c=4-4a-2b.

$$c=4-4\times 6-2\left(-22\right)$$

Calculate c from c=4-4\times 6-2\left(-22\right).

$$c=24$$

The system is now solved.

$$a=6 b=-22 c=24$$