Solve {l}{b=4a}{4a+8a+c=-4}{a-4a+c=5}

$$\left\{ \begin{array} { l } { b = 4 a } \\ { 4 a + 8 a + c = - 4 } \\ { a - 4 a + c = 5 } \end{array} \right.$$

Solve for b, a, c

$$b = -\frac{12}{5} = -2\frac{2}{5} = -2.4$$

$$a=-\frac{3}{5}=-0.6$$

$$c = \frac{16}{5} = 3\frac{1}{5} = 3.2$$

Short Steps Using Substitution

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

$$a=-\frac{1}{12}c-\frac{1}{3} c=3a+5$$

Substitute -\frac{1}{12}c-\frac{1}{3} for a in the equation c=3a+5.

$$c=3\left(-\frac{1}{12}c-\frac{1}{3}\right)+5$$

Solve c=3\left(-\frac{1}{12}c-\frac{1}{3}\right)+5 for c.

$$c=\frac{16}{5}$$

Substitute \frac{16}{5} for c in the equation a=-\frac{1}{12}c-\frac{1}{3}.

$$a=-\frac{1}{12}\times \frac{16}{5}-\frac{1}{3}$$

Calculate a from a=-\frac{1}{12}\times \frac{16}{5}-\frac{1}{3}.

$$a=-\frac{3}{5}$$

Substitute -\frac{3}{5} for a in the equation b=4a.

$$b=4\left(-\frac{3}{5}\right)$$

Calculate b from b=4\left(-\frac{3}{5}\right).

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

The system is now solved.

$$b=-\frac{12}{5} a=-\frac{3}{5} c=\frac{16}{5}$$