
Problem 9b. 1 Given the system:

$$
\left\lvert\, \begin{aligned}
\frac{1}{x+y}+x & =a-1 \\
\frac{x}{x+y} & =a-2
\end{aligned}\right.
$$

where $a$ is a real parameter.

a) Solve the system if $a=0$.

b) Find all values of $a$, such that the system has an unique solution.

c) If $a \in(2,3)$ and $(x, y)$ is a solution of the system find all values of $a$ such that the expression $\frac{x}{y}+\frac{y}{x}$ takes its minimal value.


Solution: It follows easily that $x$ and $\frac{1}{x+y}$ are roots of the equation $t^{2}-(a-1) t+a-2=0$. There are two cases to be considered:

(1) $\left|\begin{array}{cc}x & =1 \\ \frac{1}{x+y} & =a-2\end{array} \Longleftrightarrow\right| \begin{aligned} & x=1 \\ & y=\frac{3-a}{a-2},\end{aligned}$ when $a \neq 2$

and

$$
\left|\begin{array}{rl}
x & =a-2 \\
\frac{1}{x+y} & =1
\end{array} \Leftrightarrow\right| \begin{aligned}
& x=a-2 \\
& y=3-a
\end{aligned}
$$

a) When $a=0$ we have $(x, y)=\left(1 ;-\frac{3}{2}\right)$ or $(x, y)=(-2 ; 3)$.
b) The sytem has an unique solution $(0,1)$ when $a=2$ and $(1,0)$ when $a=3$.

c) If $(x, y)$ is a solution and $a \in(2 ; 3)$ then $\frac{x}{y}$ and $\frac{y}{x}$ are positive. Further $\frac{x}{y}+\frac{y}{x} \geq 2$, and equality occures when $\frac{x}{y}=\frac{y}{x}$. It follows from (1) and (2) that $\frac{x}{y}=\frac{a-2}{3-a}$. Using the equality $\frac{a-2}{3-a}=\frac{3-a}{a-2}$ we find the only value of $a=\frac{5}{2}$.
