Solve {r}{A+3B-4C=2}{-A+4B+C=8}{6A+2B+2C=4}

$$\left. \begin{array} { r } { A + 3 B - 4 C = 2 } \\ { - A + 4 B + C = 8 } \\ { 6 A + 2 B + 2 C = 4 } \end{array} \right.$$

Solve for A, B, C

$$A=-\frac{12}{67}\approx -0.179104478$$

$$B = \frac{118}{67} = 1\frac{51}{67} \approx 1.76119403$$

$$C=\frac{52}{67}\approx 0.776119403$$

Short Steps Using Substitution

Solve A+3B-4C=2 for A.

$$A=-3B+4C+2$$

Substitute -3B+4C+2 for A in the second and third equation.

$$-\left(-3B+4C+2\right)+4B+C=8 6\left(-3B+4C+2\right)+2B+2C=4$$

Solve these equations for B and C respectively.

$$B=\frac{3}{7}C+\frac{10}{7} C=-\frac{4}{13}+\frac{8}{13}B$$

Substitute \frac{3}{7}C+\frac{10}{7} for B in the equation C=-\frac{4}{13}+\frac{8}{13}B.

$$C=-\frac{4}{13}+\frac{8}{13}\left(\frac{3}{7}C+\frac{10}{7}\right)$$

Solve C=-\frac{4}{13}+\frac{8}{13}\left(\frac{3}{7}C+\frac{10}{7}\right) for C.

$$C=\frac{52}{67}$$

Substitute \frac{52}{67} for C in the equation B=\frac{3}{7}C+\frac{10}{7}.

$$B=\frac{3}{7}\times \frac{52}{67}+\frac{10}{7}$$

Calculate B from B=\frac{3}{7}\times \frac{52}{67}+\frac{10}{7}.

$$B=\frac{118}{67}$$

Substitute \frac{118}{67} for B and \frac{52}{67} for C in the equation A=-3B+4C+2.

$$A=-3\times \frac{118}{67}+4\times \frac{52}{67}+2$$

Calculate A from A=-3\times \frac{118}{67}+4\times \frac{52}{67}+2.

$$A=-\frac{12}{67}$$

The system is now solved.

$$A=-\frac{12}{67} B=\frac{118}{67} C=\frac{52}{67}$$