Schools under IB in India

 Find x, if

$\displaystyle \begin{array}{|l}4^{\frac{x}{y}+\frac{y}{x}}=32\\ \log_3(x-y)+\log_3(x+y)=1\end{array}$

Solution:
Checking if the system is defined for two variables is a hard task, so we shall find the eventual solutions to the system and check directly if the system is defined for them. We shall only write $\displaystyle \begin{array}{|l}x+y>0\\x-y>0\end{array}$ for now.
$\displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=log_432\\log_3(x^2-y^2)=1\end{array}$
$\displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}log_232\\x^2-y^2=3\end{array}$
$\displaystyle \begin{array}{|l}\frac{x^2+y^2}{xy}=\frac{5}{2}\\x^2-y^2=3\end{array}$
$\displaystyle \begin{array}{|l}x^2+y^2=\frac{5}{2}xy\\x^2-y^2=3\end{array}$
$\displaystyle \begin{array}{|l}x^2-2xy+y^2=\frac{1}{2}xy\\(x-y)(x+y)=3\end{array}$
$\displaystyle \begin{array}{|l}(x-y)^2=\frac{xy}{2}\\(x-y)(x+y)=3\end{array}$, we divide the first with the square of the second:
$\displaystyle \frac{1}{(x+y)^2}=\frac{xy}{18}$, or $\displaystyle (x+y)^2=\frac{18}{xy}$. We get the system
$\displaystyle \begin{array}{|l}x^2+2xy+y^2=\frac{18}{xy}\\x^2-2xy+y^2=\frac{xy}{2}\end{array}$, we now subtract them
$\displaystyle 4xy=\frac{18}{xy}-\frac{xy}{2}$
$\displaystyle \frac{9}{2}xy=\frac{18}{xy}$
$\displaystyle (xy)^2=4$
$\displaystyle xy = \pm 2$. But it can\'t be negative, because $\displaystyle 2(x-y)^2=xy$, therefore $\displaystyle xy=2$. Substituting back into the system, we get
$\displaystyle \begin{array}{|l}(x-y)^2=x^2-2xy+y^2=x^2+y^2-4=1\\x^2-y^2=3\end{array}$
$\displaystyle \begin{array}{|l}x^2+y^2=5\\x^2-y^2=3\end{array}$, we subtract them:
$\displaystyle 2x^2=8$, so $\displaystyle x_1=2$ and $\displaystyle x_2=-2$. $\displaystyle y_1=\frac{2}{x_1}=1$ and $\displaystyle y_2=\frac{2}{x_2}=-1$. We now have to check if they are solutions. $\displaystyle x_1+y_1=3>0$, $\displaystyle x_1-y_1=1>0$, so $\displaystyle (2;1)$ is a solution to the system. $\displaystyle x_2+y_2=-3<0$, so $\displaystyle (-2;-1)$ is not a solution.
The final solution is x=2

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