## Task A-2.1.

Determine all pairs of real numbers $(a, b)$ that satisfy the system:

$$
\begin{gathered}
a^{2}+b^{2}=25 \\
3(a+b)-a b=15 .
\end{gathered}
$$
## Solution.

From the second equation, we have $15 + ab = 3(a + b)$.

Squaring this expression, we get $225 + 30ab + a^2b^2 = 9(a + b)^2 = 9(a^2 + b^2 + 2ab). \quad 1$ point

Substituting the first equation, we get $225 + 30ab + a^2b^2 = 9(25 + 2ab). \quad 1$ point

Simplifying, we get $a^2b^2 + 12ab = 0$, or $ab(ab + 12) = 0$.

2 points

We consider two cases. If in the first case $ab = 0$, one of the numbers $a$ and $b$ is equal to zero. Substituting into the second equation, we get that the other number is equal to 5.

If in the second case $ab + 12 = 0$, or $ab = -12$, substituting into the second equation, we have $3(a + b) - (-12) = 15$, or $a + b = 1$.

1 point

Substituting $b = 1 - a$ into the first equation, we get $25 = a^2 + (1 - a)^2 = 2a^2 - 2a + 1$. Solving the corresponding quadratic equation, we get $a = 4$ or $a = -3$.

If $a = 4$, then $b = 1 - a = -3$, and for $a = -3$, we have $b = 1 - a = 4$. 1 point

Thus, the set of solutions is $\{(0,5), (5,0), (4,-3), (-3,4)\}$.

Note: A solution in which not all four solutions are found is worth a maximum of 9 points.

A part of the solution in which all four solutions are found (without proving that these are all the solutions) is worth 1 point, which corresponds to the last point in the scoring scheme.