Solve {l}{1/3a-2b+2c=2}{3/3a+6b+4c=1}{a+6b+5c=1}

$$\left. \begin{array} { l } { \frac { 1 } { 3 } a - 2 b + 2 c = 2 } \\ { \frac { 3 } { 3 } a + 6 b + 4 c = 1 } \\ { a + 6 b + 5 c = 1 } \end{array} \right.$$

Solve for a, b, c

$$a = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

$$b=-\frac{5}{12}\approx -0.416666667$$

$$c=0$$

Short Steps Using Substitution

Multiply each equation by the least common multiple of denominators in it. Simplify.

$$a-6b+6c=6 3a+18b+12c=3 a+6b+5c=1$$

Solve a-6b+6c=6 for a.

$$a=6+6b-6c$$

Substitute 6+6b-6c for a in the second and third equation.

$$3\left(6+6b-6c\right)+18b+12c=3 6+6b-6c+6b+5c=1$$

Solve these equations for b and c respectively.

$$b=-\frac{5}{12}+\frac{1}{6}c c=5+12b$$

Substitute -\frac{5}{12}+\frac{1}{6}c for b in the equation c=5+12b.

$$c=5+12\left(-\frac{5}{12}+\frac{1}{6}c\right)$$

Solve c=5+12\left(-\frac{5}{12}+\frac{1}{6}c\right) for c.

$$c=0$$

Substitute 0 for c in the equation b=-\frac{5}{12}+\frac{1}{6}c.

$$b=-\frac{5}{12}+\frac{1}{6}\times 0$$

Calculate b from b=-\frac{5}{12}+\frac{1}{6}\times 0.

$$b=-\frac{5}{12}$$

Substitute -\frac{5}{12} for b and 0 for c in the equation a=6+6b-6c.

$$a=6+6\left(-\frac{5}{12}\right)-6\times 0$$

Calculate a from a=6+6\left(-\frac{5}{12}\right)-6\times 0.

$$a=\frac{7}{2}$$

The system is now solved.

$$a=\frac{7}{2} b=-\frac{5}{12} c=0$$