Solve {l}{M+C+H=10}{M+4C+4H=32}{10M+18C+10H=78}

$$\left. \begin{array} { l } { M + C + H = 10 } \\ { M + 4 C + 4 H = 32 } \\ { 10 M + 18 C + 10 H = 78 } \end{array} \right.$$

Solve for M, C, H

$$M = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

$$C = -\frac{11}{4} = -2\frac{3}{4} = -2.75$$

$$H = \frac{121}{12} = 10\frac{1}{12} \approx 10.083333333$$

Short Steps Using Substitution

Solve M+C+H=10 for M.

$$M=-C-H+10$$

Substitute -C-H+10 for M in the second and third equation.

$$-C-H+10+4C+4H=32 10\left(-C-H+10\right)+18C+10H=78$$

Solve these equations for H and C respectively.

$$H=-C+\frac{22}{3} C=-\frac{11}{4}$$

Substitute -\frac{11}{4} for C in the equation H=-C+\frac{22}{3}.

$$H=-\left(-\frac{11}{4}\right)+\frac{22}{3}$$

Calculate H from H=-\left(-\frac{11}{4}\right)+\frac{22}{3}.

$$H=\frac{121}{12}$$

Substitute \frac{121}{12} for H and -\frac{11}{4} for C in the equation M=-C-H+10.

$$M=-\left(-\frac{11}{4}\right)-\frac{121}{12}+10$$

Calculate M from M=-\left(-\frac{11}{4}\right)-\frac{121}{12}+10.

$$M=\frac{8}{3}$$

The system is now solved.

$$M=\frac{8}{3} C=-\frac{11}{4} H=\frac{121}{12}$$