LVIII OM - I - Problem 1

Problem 1.
Solve in real numbers $ x $, $ y $, $ z $ the system of equations
Subtracting the second equation of the system from the first, we get






Similarly, by subtracting the third equation from the first, we obtain the equation $ 0=(x-z)(x+z-2y+5) $. Therefore, the system of equations given in the problem is equivalent to the system



The product of two numbers is equal to zero if and only if at least one of the factors is equal to zero. Therefore, system (1) is equivalent to the following alternative of four systems: 












In the first of the systems (2), the second and third equations lead to the conclusion that $ x=y=z $. Using this relationship in the first equation of this system, we obtain the quadratic equation $ 3x^2+5x=2 $, which has two solutions: $ x=-2 $ and $ x={1\over 3} $. Thus, the considered system of equations has two solutions: $ (-2,-2,-2) $ and $ ({1\over 3},{1\over 3},{1\over 3}) $. 
Now let's examine the second system (2). The second equation of this system means that $ x=y $; combining this with the third equation, we see that $ z=-x+2y-5=x-5 $. Therefore, the first equation of the system takes the form


$$2=x^2+2yz+5x=x^2+2x(x-5)+5x=3x^2-5x.$$



The quadratic equation $ 3x^2-5x=2 $ has solutions $ x=2 $ and $ x=-{1\over 3} $. This gives two solutions of the system: $ (2,2,-3) $ and $ (-{1\over 3},-{1\over 3},-{16\over 3}) $. 
Similarly, we solve the third system (2), obtaining the solutions $ (2,-3,2) $ and $ (-{1\over 3},-{16\over 3},-{1\over 3}) $. 
Finally, let's move on to the last system (2). Combining the second and third equations, we see that $ y-2z=z-2y $, which means $ y=z $. Therefore, the second equation takes the form $ y=x+5 $, and the first can be written as


$$2=x^2+2yz+5x=(y-5)^2+2y^2+5(y-5)=3y^2-5y,$$



which, after solving the quadratic equation, gives $ y=2 $ and $ y=-{1\over 3} $. As a result, we get two solutions: $ (-3,2,2) $ and $ (-{16\over 3},-{1\over 3},-{1\over 3}) $. 
Finally, it remains to verify that each of the 8 triples indeed represents a solution to the system of equations given in the problem. 
Answer: The solutions $ (x, y, z) $ of the given system of equations are: