Solve the system
$$x+y+\dfrac{1}{x}+\dfrac{1}{y} = \dfrac{9}{2} \
xy+\dfrac{1}{xy} =\dfrac{5}{2} $$

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Uniteasy Admin · Accepted Answer

Let 
$$a=xy$$
$$b=x+y$$
The two equations are transformed to
$$b+\dfrac{b}{a} = \dfrac{9}{2} $$n
$$a+\dfrac{1}{a} =\dfrac{5}{2} $$n
Solve (2) for $$a$$
$$a^2-\dfrac{5}{2}a+1=0 $$
Then we get
$$a=\dfrac{1}{2}$$ or $$a=2 $$
Substituting the value of $$a$$ to (1) gives
$$b=\dfrac{3}{2}$$ or $$b = 3 $$
Then we have two cases for $$x,y$$
$$\begin{cases} xy=\dfrac{1}{2} \ x+y=\dfrac{3}{2} \end{cases}$$n
and
$$\begin{cases} xy=2 \ x+y = 3 \end{cases}$$n
Case 1, $$x,y$$ is the roots for the quadratic equation $$z^2-\dfrac{3}{2}z+\dfrac{1}{2} =0$$ based on Vieta's formula.
Solving the equation gives two solutions for $$x,y$$
$$\begin{cases} x=\dfrac{1}{2} \ y=1 \end{cases} \quad	ext{or}\quad \begin{cases}  x=1 \ y=\dfrac{1}{2} \end{cases} $$
Case 2, $$x,y$$ is the roots for the quadratic equation $$z^2-3z+2 =0$$ based on Vieta's formula. Then we get another 2 solutions.
$$\begin{cases} x=1\ y=2 \end{cases} \quad	ext{or}\quad \begin{cases}  x=2 \ y=1 \end{cases} $$
In summary, there're four solutions for the system of equations.

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