Solve {c}{a+b+c=107}{c=4b}{b=5-a}

$$\left. \begin{array} { c } { a + b + c = 107 } \\ { c = 4 b } \\ { b = 5 - a } \end{array} \right.$$

Solve for a, b, c

$$a = -\frac{41}{2} = -20\frac{1}{2} = -20.5$$

$$b = \frac{51}{2} = 25\frac{1}{2} = 25.5$$

$$c=102$$

Short Steps Using Substitution

Reorder the equations.

$$c=4b a+b+c=107 b=5-a$$

Substitute 4b for c in the equation a+b+c=107.

$$a+b+4b=107$$

Solve the second equation for b and the third equation for a.

$$b=-\frac{1}{5}a+\frac{107}{5} a=-b+5$$

Substitute -\frac{1}{5}a+\frac{107}{5} for b in the equation a=-b+5.

$$a=-\left(-\frac{1}{5}a+\frac{107}{5}\right)+5$$

Solve a=-\left(-\frac{1}{5}a+\frac{107}{5}\right)+5 for a.

$$a=-\frac{41}{2}$$

Substitute -\frac{41}{2} for a in the equation b=-\frac{1}{5}a+\frac{107}{5}.

$$b=-\frac{1}{5}\left(-\frac{41}{2}\right)+\frac{107}{5}$$

Calculate b from b=-\frac{1}{5}\left(-\frac{41}{2}\right)+\frac{107}{5}.

$$b=\frac{51}{2}$$

Substitute \frac{51}{2} for b in the equation c=4b.

$$c=4\times \frac{51}{2}$$

Calculate c from c=4\times \frac{51}{2}.

$$c=102$$

The system is now solved.

$$a=-\frac{41}{2} b=\frac{51}{2} c=102$$