Solve a system by putting new variables

Question

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 $$\begin{array}{|l} (x+2y)^2=a \\ (y-2x)^2=b \end{array}...$$

Answer 1

Rewrite the first equation as $$ \frac{x^2}{xy}+\frac{y^2}{xy}=\frac{7}{25}. $$ The second equation then gives $\frac{625}{7}=xy$. Substituting gives $$ 49x^4 - 1225x^2 + 390625=0, $$ which is a biquadratic equation with no real solution. This seems short enough.

Answer 2

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