Solve {l}{3b+32+4a=a}{4c-3b+2a=b}{8a+b+b=c}

$$\left. \begin{array} { l } { 3 b + 32 + 4 a = a } \\ { 4 c - 3 b + 2 a = b } \\ { 8 a + b + b = c } \end{array} \right.$$

Solve for b, a, c

$$b = -\frac{544}{45} = -12\frac{4}{45} \approx -12.088888889$$

$$a = \frac{64}{45} = 1\frac{19}{45} \approx 1.422222222$$

$$c = -\frac{64}{5} = -12\frac{4}{5} = -12.8$$

Short Steps Using Substitution

Reorder the equations.

$$8a+b+b=c 4c-3b+2a=b 3b+32+4a=a$$

Solve 8a+b+b=c for c.

$$c=8a+2b$$

Substitute 8a+2b for c in the equation 4c-3b+2a=b.

$$4\left(8a+2b\right)-3b+2a=b$$

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

$$a=-\frac{2}{17}b b=-\frac{32}{3}-a$$

Substitute -\frac{2}{17}b for a in the equation b=-\frac{32}{3}-a.

$$b=-\frac{32}{3}-\left(-\frac{2}{17}b\right)$$

Solve b=-\frac{32}{3}-\left(-\frac{2}{17}b\right) for b.

$$b=-\frac{544}{45}$$

Substitute -\frac{544}{45} for b in the equation a=-\frac{2}{17}b.

$$a=-\frac{2}{17}\left(-\frac{544}{45}\right)$$

Calculate a from a=-\frac{2}{17}\left(-\frac{544}{45}\right).

$$a=\frac{64}{45}$$

Substitute \frac{64}{45} for a and -\frac{544}{45} for b in the equation c=8a+2b.

$$c=8\times \frac{64}{45}+2\left(-\frac{544}{45}\right)$$

Calculate c from c=8\times \frac{64}{45}+2\left(-\frac{544}{45}\right).

$$c=-\frac{64}{5}$$

The system is now solved.

$$b=-\frac{544}{45} a=\frac{64}{45} c=-\frac{64}{5}$$