حل {l}{NI+C+H=TU}{2M+4C+4H=32}{10M+18C+10H=78}

$$\left. \begin{array} { l } { NI + C + H = TU } \\ { 2 M + 4 C + 4 H = 32 } \\ { 10 M + 18 C + 10 H = 78 } \end{array} \right.$$

حل مسائل N، I، C، H، T، U، M

$$N=-\frac{C+H-TU}{I}\text{, }I\neq 0\text{, }C=41-5H\text{, }H\in \mathrm{R}\text{, }T\in \mathrm{R}\text{, }U\in \mathrm{R}\text{, }M=2\left(4H-33\right)$$

$$N\in \mathrm{R}\text{, }I=0\text{, }C=41-5H\text{, }H\in \mathrm{R}\text{, }T\neq 0\text{, }U=-\frac{4H-41}{T}\text{, }M=2\left(4H-33\right)$$

$$N\in \mathrm{R}\text{, }I=0\text{, }C=-\frac{41}{4}=-10.25\text{, }H=\frac{41}{4}=10.25\text{, }T=0\text{, }U\in \mathrm{R}\text{, }M=16$$