Find all pairs of real numbers $x$ and $y$ which satisfy the following equations: \begin{align*} x^2 + y^2 - 48x - 29y + 714 & = 0 \\ 2xy - 29x - 48y + 756 & = 0 \end{align*}
Problem
Source: South African Mathematics Olympiad 2022, Problem 2
Tags: quadratics, algebra, system of equations
12.08.2022 01:11
Check this link for the solution
12.08.2022 02:00
Sorry for adding a duplicate post of the problem. I searched to see if anyone else had added the problems, but only managed to find question 3 and question 6. I see now that you did in fact add all of them.
12.08.2022 02:31
I am not sure I added P4, but I added others
20.09.2022 11:25
Add the two equations together, we have $(x+y-\dfrac {77} 2)^2=\dfrac {49} 4$ so $x+y-\dfrac {77} 2=\pm \dfrac 7 2$ Subtract the second equation from the first oneļ¼ we have $(x-y-\dfrac {19} 2)^2=\dfrac {529} 4$ so $x-y-\dfrac {19} 2=\pm \dfrac {23} 2$ as we have $2 \times 2 \qquad \pm$, we have 4 answers: $$ \begin{cases} x=31.5 \\ y=10.5 \end{cases} \qquad \begin{cases} x=20 \\ y=22 \end{cases}$$$$ \begin{cases} x=28 \\ y=7 \end{cases}\qquad \begin{cases} x=16.5 \\ y=18.5 \end{cases}$$
31.08.2023 15:30
When we add both eqautions, we get X^2 + y^2 + 2xy - 77x -77y + 1470 = 0 (X + Y) ^2 -77(X + Y) + 1470 = 0 (X + Y) ^2 -35(X +Y) - 42 (X + Y) + 1470 = 0 (X + Y) (X + Y - 35 ) - 42 (X+ Y - 35) = 0 (X + Y - 42) ( X + Y - 35) = 0 So, either x +y = 35 or 42 And when we subtract them, we get x - y = 21 or -2 This will result in 4 solutions i.e. (28, 7); (31.5, 10.5); (20, 22); (16.5, 18.5)