Solve {l}{16a+4b+c=2}{4a+2b+c=4}{36a+6b+c=1}

$$\left. \begin{array} { l } { 16 a + 4 b + c = 2 } \\ { 4 a + 2 b + c = 4 } \\ { 36 a + 6 b + c = 1 } \end{array} \right.$$

Solve for a, b, c

$$a=\frac{1}{8}=0.125$$

$$b = -\frac{7}{4} = -1\frac{3}{4} = -1.75$$

$$c=7$$

Short Steps Using Substitution

Solve 16a+4b+c=2 for c.

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

Substitute 2-16a-4b for c in the second and third equation.

$$4a+2b+2-16a-4b=4 36a+6b+2-16a-4b=1$$

Solve these equations for b and a respectively.

$$b=-1-6a a=-\frac{1}{10}b-\frac{1}{20}$$

Substitute -1-6a for b in the equation a=-\frac{1}{10}b-\frac{1}{20}.

$$a=-\frac{1}{10}\left(-1-6a\right)-\frac{1}{20}$$

Solve a=-\frac{1}{10}\left(-1-6a\right)-\frac{1}{20} for a.

$$a=\frac{1}{8}$$

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

$$b=-1-6\times \frac{1}{8}$$

Calculate b from b=-1-6\times \frac{1}{8}.

$$b=-\frac{7}{4}$$

Substitute -\frac{7}{4} for b and \frac{1}{8} for a in the equation c=2-16a-4b.

$$c=2-16\times \frac{1}{8}-4\left(-\frac{7}{4}\right)$$

Calculate c from c=2-16\times \frac{1}{8}-4\left(-\frac{7}{4}\right).

$$c=7$$

The system is now solved.

$$a=\frac{1}{8} b=-\frac{7}{4} c=7$$