[tex]\left \{ {{x-2y+1=0} \atop {}\sqrt{x}+ \sqrt{y}=2 } \right.[/tex]

Предыдущий
вопрос Следующий
вопрос

To solve this system of equations, we can start by isolating one variable in the first equation and then substituting it into the second equation. Let's solve for x in the first equation:

[x - 2y + 1 = 0]
[x = 2y - 1]

Now, substitute this expression for x into the second equation:

[\sqrt{2y - 1} + \sqrt{y} = 2]

Square both sides to eliminate the square roots:

[(\sqrt{2y - 1} + \sqrt{y})^2 = 2^2]
[2y - 1 + 2\sqrt{y(2y - 1)} + y = 4]
[2y - 1 + 2\sqrt{2y^2 - y} + y = 4]
[3y + 2\sqrt{2y^2 - y} = 5]
[2\sqrt{2y^2 - y} = 5 - 3y]
[4(2y^2 - y) = (5 - 3y)^2]
[8y^2 - 4y = 25 - 30y + 9y^2]
[8y^2 - 4y - 9y^2 + 30y - 25 = 0]
[-y^2 + 26y - 25 = 0]
[y^2 - 26y + 25 = 0]
[(y - 1)(y - 25) = 0]

Therefore, y = 1 or y = 25.

Using these values to solve for x using the equation x = 2y - 1, we get:

If y = 1, then x = 2(1) - 1 = 1
If y = 25, then x = 2(25) - 1 = 49

Therefore, the solutions to the system of equations are:
(1, 1) and (49, 25)