B2. Find all pairs of real numbers $x$ and $y$ that solve the system of equations

$$
\begin{gathered}
\sqrt{x}-\sqrt{y}=1 \\
\sqrt{8 x+7}-\sqrt{8 y+7}=2
\end{gathered}
$$

## Problems for 3rd Year

Time for solving: 45 minutes.
B2. We transform the equations into equivalent equations $\sqrt{x}=1+\sqrt{y}$ and $\sqrt{8 x+7}=2+\sqrt{8 y+7}$ and then square them to get

$$
\begin{aligned}
x & =1+2 \sqrt{y}+y \quad \text { and } \\
8 x+7 & =4+4 \sqrt{8 y+7}+8 y+7
\end{aligned}
$$

We subtract 8 times the first equation from the second equation and simplify the resulting equation to get $1+4 \sqrt{y}=\sqrt{8 y+7}$. We square this equation and simplify it to $4 \sqrt{y}=$ $3-4 y$. When we square the equation again, we get the quadratic equation $16 y^{2}-40 y+9=0$, which can be factored as $(4 y-1)(4 y-9)=0$. This gives us two solutions, $y_{1}=\frac{1}{4}$ and $y_{2}=\frac{9}{4}$. Using the equation $x=1+2 \sqrt{y}+y$, we calculate $x_{1}=\frac{9}{4}$ and $x_{2}=\frac{25}{4}$. We can verify that the pair $\left(x_{1}, y_{1}\right)$ satisfies the original system of equations, while the pair $\left(x_{2}, y_{2}\right)$ does not, since

$$
\sqrt{8 \cdot \frac{25}{4}+7}-\sqrt{8 \cdot \frac{9}{4}+7}=\sqrt{57}-5
$$

which is not an integer and therefore cannot be equal to 2. Thus, the only solution to the problem is the pair $x=\frac{9}{4}$ and $y=\frac{1}{4}$.

Transforming the equations into equivalent equations $\sqrt{x}=1+\sqrt{y}$ and $\sqrt{8 x+7}=2+\sqrt{8 y+7} \ldots \mathbf{. 2}$ points

Squaring the equations 3 points

Writing the equation in one unknown 5 points

Squaring the equation $1+4 \sqrt{y}=\sqrt{8 y+7}$ 2 points

Writing the quadratic equation $16 y^{2}-40 y+9=0$ 2 points

Calculating the solutions $y_{1}$ and $y_{2}$ 2 points

Calculating $x_{1}$ and $x_{2}$ 2 points

Excluding the solution $\left(\frac{25}{4}, \frac{9}{4}\right)$

2 points

## 63rd Mathematical Competition for High School Students in Slovenia

Selection Competition, March 21, 2019

## Solutions for 3rd Year