276. Find $x$ and $y$ from the system of equations

$$
\frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=a, \quad \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=b
$$

($a$ and $b$ are given numbers).
276. Answer: $x=\frac{a+b \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}} ; y=\frac{b+a \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}}$.

By adding and subtracting the given equations, and then multiplying the results, we obtain the following consequence of the system:

$$
x^{2}-y^{2}=a^{2}-b^{2}, \quad \text { i.e. } \sqrt{x^{2}-y^{2}}=\sqrt{a^{2}-b^{2}}
$$

Substituting this into the original system and solving it (as a linear system), we find the answer. The fact that the obtained formulas coincide with the original ones with the substitution $a$ for $x$, $b$ for $-y$, allows us to skip the verification: from the formulas of the answer, we will get, as a consequence, similar formulas with the substitution $x$ for $a$, $y$ for $-b$, i.e., the original system.

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-238.jpg?height=338&width=329&top_left_y=397&top_left_x=164)

Q

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-238.jpg?height=458&width=370&top_left_y=333&top_left_x=608)

Fig. 113

$\nabla$ Note that nothing would change if we considered the following system instead of the given one:

$$
\frac{x-y f\left(x^{2}-y^{2}\right)}{\sqrt{1-f^{2}\left(x^{2}-y^{2}\right)}}=a, \frac{y-x f\left(x^{2}-y^{2}\right)}{\sqrt{1-f^{2}\left(x^{2}-y^{2}\right)}}=b
$$

where $f$ is any function not exceeding 1 in absolute value. The essence of this problem is that the "Lorentz transformation"

$$
\begin{gathered}
(x, y) \rightarrow\left(x^{\prime}, y^{\prime}\right), \quad x^{\prime}=\frac{x-v y}{\sqrt{1-v^{2}}} \\
y^{\prime}=\frac{y-x v}{\sqrt{1-v^{2}}}
\end{gathered}
$$

which plays a fundamental role in Einstein's theory of relativity, preserves the quantity $x^{2}-y^{2}: x^{2}-y^{2}=$ $=\left(x^{\prime}\right)^{2}-\left(y^{\prime}\right)^{2}$.