求解 8/x+2y+3/2x-y=3

$$\frac{ 8 }{ x+2y } + \frac{ 3 }{ 2x-y } =3$$

求解 y 的值 (复数求解)

$$\left\{\begin{matrix}\\y=\frac{\sqrt{225x^{2}-420x+4}}{12}+\frac{3x}{4}+\frac{1}{6}\text{, }&\text{unconditionally}\\y=-\frac{\sqrt{225x^{2}-420x+4}}{12}+\frac{3x}{4}+\frac{1}{6}\text{, }&x\neq 0\end{matrix}\right.$$

求解 x 的值

$$\left\{\begin{matrix}\\x=\frac{\sqrt{225y^{2}-390y+361}}{12}-\frac{3y}{4}+\frac{19}{12}\text{, }&\text{unconditionally}\\x=-\frac{\sqrt{225y^{2}-390y+361}}{12}-\frac{3y}{4}+\frac{19}{12}\text{, }&y\neq 0\end{matrix}\right.$$

求解 y 的值

$$\left\{\begin{matrix}y=-\frac{\sqrt{225x^{2}-420x+4}}{12}+\frac{3x}{4}+\frac{1}{6}\text{, }&x\geq \frac{8\sqrt{3}+14}{15}\text{ or }\left(x\neq 0\text{ and }x\leq \frac{14-8\sqrt{3}}{15}\right)\\y=\frac{\sqrt{225x^{2}-420x+4}}{12}+\frac{3x}{4}+\frac{1}{6}\text{, }&x\geq \frac{8\sqrt{3}+14}{15}\text{ or }x\leq \frac{14-8\sqrt{3}}{15}\end{matrix}\right.$$