Solve {c}{20a-10b-8c=33}{-10a+24b-4c=12.2}{-8a-4b+20c=-12.2}

$$\left\{ \begin{array} { c } { 20 a - 10 b - 8 c = 33 } \\ { - 10 a + 24 b - 4 c = 12.2 } \\ { - 8 a - 4 b + 20 c = - 12.2 } \end{array} \right.$$

Solve for a, b, c

$$a=3$$

$$b = \frac{557}{290} = 1\frac{267}{290} \approx 1.920689655$$

$$c=\frac{113}{116}\approx 0.974137931$$

Short Steps Using Substitution

Solve 20a-10b-8c=33 for a.

$$a=\frac{1}{2}b+\frac{2}{5}c+\frac{33}{20}$$

Substitute \frac{1}{2}b+\frac{2}{5}c+\frac{33}{20} for a in the second and third equation.

$$-10\left(\frac{1}{2}b+\frac{2}{5}c+\frac{33}{20}\right)+24b-4c=12.2 -8\left(\frac{1}{2}b+\frac{2}{5}c+\frac{33}{20}\right)-4b+20c=-12.2$$

Solve these equations for b and c respectively.

$$b=\frac{287}{190}+\frac{8}{19}c c=\frac{5}{84}+\frac{10}{21}b$$

Substitute \frac{287}{190}+\frac{8}{19}c for b in the equation c=\frac{5}{84}+\frac{10}{21}b.

$$c=\frac{5}{84}+\frac{10}{21}\left(\frac{287}{190}+\frac{8}{19}c\right)$$

Solve c=\frac{5}{84}+\frac{10}{21}\left(\frac{287}{190}+\frac{8}{19}c\right) for c.

$$c=\frac{113}{116}$$

Substitute \frac{113}{116} for c in the equation b=\frac{287}{190}+\frac{8}{19}c.

$$b=\frac{287}{190}+\frac{8}{19}\times \frac{113}{116}$$

Calculate b from b=\frac{287}{190}+\frac{8}{19}\times \frac{113}{116}.

$$b=\frac{557}{290}$$

Substitute \frac{557}{290} for b and \frac{113}{116} for c in the equation a=\frac{1}{2}b+\frac{2}{5}c+\frac{33}{20}.

$$a=\frac{1}{2}\times \frac{557}{290}+\frac{2}{5}\times \frac{113}{116}+\frac{33}{20}$$

Calculate a from a=\frac{1}{2}\times \frac{557}{290}+\frac{2}{5}\times \frac{113}{116}+\frac{33}{20}.

$$a=3$$

The system is now solved.

$$a=3 b=\frac{557}{290} c=\frac{113}{116}$$