ដោះស្រាយ {l}{x-y=12}{xy={(frac{6*{(4)}+1}{4})}}{z=x+y}{a=z}{text{Solvefor}btext{where}}{b=a}

$$\left. \begin{array} { l } { x - y = 12 }\\ { x y = {(\frac{6 \cdot {(4)} + 1}{4})} }\\ { z = x + y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

ដោះស្រាយសម្រាប់ x, y, z, a, b

$$x=-\frac{1}{2}=-0.5\text{, }y=-\frac{25}{2}=-12.5\text{, }z=-13\text{, }a=-13\text{, }b=-13$$

$$x=\frac{25}{2}=12.5\text{, }y=\frac{1}{2}=0.5\text{, }z=13\text{, }a=13\text{, }b=13$$