{l}{{(3x+1)}{(x+2)}=h{(6+x)}}{y=t}{z=y}{a=z}{text{Solvefor}btext{where}}{b=a} ھەل قىلىش

$$\left. \begin{array} { l } { {(3 x + 1)} {(x + 2)} = h {(6 + x)} }\\ { y = t }\\ { z = y }\\ { a = z }\\ { \text{Solve for } b \text{ where} } \\ { b = a } \end{array} \right.$$

x، y، z نى يېشىش (complex solution)

$$x=\frac{-\sqrt{h^{2}+58h+25}+h-7}{6}\text{, }y=a\text{, }z=a$$

$$x=\frac{\sqrt{h^{2}+58h+25}+h-7}{6}\text{, }y=a\text{, }z=a\text{, }b=a\text{ and }t=a$$

x، y، z نى يېشىش

$$x=\frac{-\sqrt{h^{2}+58h+25}+h-7}{6}\text{, }y=t\text{, }z=t$$

$$x=\frac{\sqrt{h^{2}+58h+25}+h-7}{6}\text{, }y=t\text{, }z=t\text{, }a=b\text{ and }t=b\text{ and }\left(h\geq 4\sqrt{51}-29\text{ or }h\leq -4\sqrt{51}-29\right)$$