Solve {l}{4=A+B+C}{-2=-A-3B+2C}{-12=-6A}

$$\left. \begin{array} { l } { 4 = A + B + C } \\ { - 2 = - A - 3 B + 2 C } \\ { - 12 = - 6 A } \end{array} \right.$$

Solve for A, B, C

$$A=2$$

$$B=\frac{4}{5}=0.8$$

$$C = \frac{6}{5} = 1\frac{1}{5} = 1.2$$

Short Steps Using Substitution

Solve 4=A+B+C for A.

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

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

$$-2=-\left(-B+4-C\right)-3B+2C -12=-6\left(-B+4-C\right)$$

Solve these equations for B and C respectively.

$$B=-1+\frac{3}{2}C C=2-B$$

Substitute -1+\frac{3}{2}C for B in the equation C=2-B.

$$C=2-\left(-1+\frac{3}{2}C\right)$$

Solve C=2-\left(-1+\frac{3}{2}C\right) for C.

$$C=\frac{6}{5}$$

Substitute \frac{6}{5} for C in the equation B=-1+\frac{3}{2}C.

$$B=-1+\frac{3}{2}\times \frac{6}{5}$$

Calculate B from B=-1+\frac{3}{2}\times \frac{6}{5}.

$$B=\frac{4}{5}$$

Substitute \frac{4}{5} for B and \frac{6}{5} for C in the equation A=-B+4-C.

$$A=-\frac{4}{5}+4-\frac{6}{5}$$

Calculate A from A=-\frac{4}{5}+4-\frac{6}{5}.

$$A=2$$

The system is now solved.

$$A=2 B=\frac{4}{5} C=\frac{6}{5}$$