Solve {l}{A+B=1/6}{B+C=1/7}{A+C=1/12}

$$\left. \begin{array} { l } { A + B = \frac { 1 } { 6 } } \\ { B + C = \frac { 1 } { 7 } } \\ { A + C = \frac { 1 } { 12 } } \end{array} \right.$$

Solve for A, B, C

$$A=\frac{3}{56}\approx 0.053571429$$

$$B=\frac{19}{168}\approx 0.113095238$$

$$C=\frac{5}{168}\approx 0.029761905$$

Short Steps Using Substitution

Solve A+B=\frac{1}{6} for A.

$$A=-B+\frac{1}{6}$$

Substitute -B+\frac{1}{6} for A in the equation A+C=\frac{1}{12}.

$$-B+\frac{1}{6}+C=\frac{1}{12}$$

Solve the second equation for B and the third equation for C.

$$B=-C+\frac{1}{7} C=B-\frac{1}{12}$$

Substitute -C+\frac{1}{7} for B in the equation C=B-\frac{1}{12}.

$$C=-C+\frac{1}{7}-\frac{1}{12}$$

Solve C=-C+\frac{1}{7}-\frac{1}{12} for C.

$$C=\frac{5}{168}$$

Substitute \frac{5}{168} for C in the equation B=-C+\frac{1}{7}.

$$B=-\frac{5}{168}+\frac{1}{7}$$

Calculate B from B=-\frac{5}{168}+\frac{1}{7}.

$$B=\frac{19}{168}$$

Substitute \frac{19}{168} for B in the equation A=-B+\frac{1}{6}.

$$A=-\frac{19}{168}+\frac{1}{6}$$

Calculate A from A=-\frac{19}{168}+\frac{1}{6}.

$$A=\frac{3}{56}$$

The system is now solved.

$$A=\frac{3}{56} B=\frac{19}{168} C=\frac{5}{168}$$