Question

(×+2)(y-1)=0 ×в квадрате -xy-12=0 Система уравнений

Answer 1

$(x+2)(y-1)=0 \\ x^2-xy-12=0 \\ x=-2 y = 1 \\ -2^2+2y-12=0 \\ 4+2y-12=0 \\ -8+2y=0 \\ -4+y=0 \\ y = 4 \\ x^2-4x-12=0 \\ D = 16-4*1*-12=64^(1/2) \\ x_1,x_2 = 4+-8/2\\ x_1 = 6, x_2 = -2 \\ (x= -2, y = 4) ( y= 1; x_1 = 6, x_2 = -2)$

Answer 2

To solve this system of equations, we can use substitution or elimination method.

Method 1: Substitution

From the first equation, we have two cases:

Case 1: x+2 = 0 If x+2=0, then x=-2. Substituting x=-2 into the second equation, we get: (-2)^2 - y(-2) - 12 = 0 4 + 2y - 12 = 0 2y = 8 y = 4 Therefore, one solution to the system is (-2,4).

Case 2: y-1 = 0 If y-1=0, then y=1. Substituting y=1 into the first equation, we get: (x+2)(1-1) = 0 x+2 = 0 x = -2 Substituting x=-2 into the second equation, we get: (-2)^2 - 1(-2) - 12 = 0 4 + 2 + 12 = 0 This equation has no real solutions, so we discard this case.

Therefore, the only solution to the system is (-2,4).

Method 2: Elimination

From the first equation, we have x+2=0 or y-1=0. If x+2=0, then x=-2. Substituting x=-2 into the second equation, we get: (-2)^2 - y(-2) - 12 = 0 4 + 2y - 12 = 0 2y = 8 y = 4 Therefore, one solution to the system is (-2,4).

If y-1=0, then y=1. Substituting y=1 into the first equation, we get: (x+2)(1-1) = 0 x+2 = 0 x = -2 Substituting x=-2 into the second equation, we get: (-2)^2 - 1(-2) - 12 = 0 4 + 2 + 12 = 0 This equation has no real solutions, so we discard this case.

Therefore, the only solution to the system is (-2,4).