Question

(x - 2)(y + 1) = 0,6y2 + x - y = 3.​

Answer 1

Ответ:

(-4;-1) ; (2; -1/3) ; (2;0,5)

Пошаговое объяснение:

Заметим, что произведение в 1-ом уравнение равно нулю => один из множителей равен нулю

Значит у нас 2 случая: когда х=2 или у=-1

При х=2 преобразуем 2 ур-ие:

6у²+2-у = 3

6у²-у-1 = 0

D = b²-4ac = 1 +4*6 = 25 =5²

y1,2 = (-b±√D)/2a = (1±5)/12 => y1= 0,5 и у2= -1/3

При у= -1 преобразуем 2 ур-ие

6(-1)²+х-(-1) = 3

6+х+1 = 3

х= -4

Answer 2

To solve the system of equations:

1. $(x - 2)(y + 1) = 0$
2. $0.6y^2 + x - y = 3$

Let's start by solving equation (1) for $x$ and $y$ separately:

1. $(x - 2)(y + 1) = 0$ The product of two factors is zero if at least one of the factors is zero. So we have two cases:

   Case 1: $x - 2 = 0$ $x = 2$

   Case 2: $y + 1 = 0$ $y = -1$

Now we substitute these values in equation (2) to find the corresponding values of $y$ and $x$ that satisfy the second equation:

2. $0.6y^2 + x - y = 3$

   For $x = 2$ and $y = -1$, we have:

   $0.6(-1)^2 + 2 - (-1) = 3$ $0.6 + 2 + 1 = 3$ $3.6 = 3$

The values of $x = 2$ and $y = -1$ don't satisfy equation (2), so there seems to be no solution that simultaneously satisfies both equations.

Hence, there might be a mistake in the given equations, or they might not be compatible with each other. Please review the equations or the problem description.