Solve a system by putting new variables 
Solve the system: $$\begin{array}{|l}  
 \dfrac{x}{y}+\dfrac{y}{x}=\dfrac{7}{25} \\  x^2+y^2=25  \end{array}$$

The first step is to determine the domain: $\begin{array}{|l} x \ne 0 \\ y \ne 0 \end{array}$
We can simplify the first equation of the system, and we get: $\begin{array}{|l} \dfrac{x^2+y^2}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}$
$\begin{array}{|l} \dfrac{25}{xy}=\dfrac{7}{25} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} 7xy=625 \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} xy=\dfrac{625}{7} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ x^2+y^2=25 \end{array}\Leftrightarrow \begin{array}{|l} x=\dfrac{625}{7y} \\ y^4-25y^2+\dfrac{390625}{49}=0 \end{array}$
The equation: $$y^4-25y^2+\dfrac{390625}{49}=0$$ has no solutions, so the whole system does not have a solution. In the text of the problem is said I should solve by "putting a new variable". I don't know how this method is called in English, and I would be grateful if you tell me. Let me give you a basic example of a system that can be solved using this method:
$$\begin{array}{|l} (x+2y)^2-(y-2x)^2=168 \\ (x+2y)^2+(y-2x)^2=12 \end{array}$$ 
Let $$\begin{array}{|l} (x+2y)^2=a \\ (y-2x)^2=b \end{array}...$$

let $b=\frac{x}{y}$ (defined because $x\ne 0,y\ne 0$). Then by the first equation:
$b+\frac{1}{b}=\frac{7}{25}$
$\iff (b\ne0)$
$b^2-\frac{7}{25}b+1=0$
$\iff$
False (no real solutions)
so there are no solutions
