6- y = 4 3- 2 y = x ^ 2 -2 . లు Which of the following shows the solution (s) for this system of equations ? point There are no solutions (- 3, 0) and (3, 0); (- 2, 4) and (2, 4); (0, 8)

To find the solution(s) for this system of equations, we need to solve them simultaneously. Let's solve them one by one:

Equation 1: 6 - y = 4
Rearranging this equation, we get:
y = 6 - 4
y = 2

So, the solution for Equation 1 is (2).

Equation 2: 3 - 2y = x^2 - 2
Rearranging this equation, we get:
x^2 = 3 - 2y + 2
x^2 = 5 - 2y
Since this equation has an x^2 term, it suggests that the solutions will not be single points but rather a set of points. Let's solve for x:

x = ±√(5 - 2y)

Now we can check the given points to find which ones are solutions:

1. (-3, 0):
x = ±√(5 - 2(0))
x = ±√5
So the solution for this point is (-3 ± √5, 0)

2. (3, 0):
x = ±√(5 - 2(0))
x = ±√5
So the solution for this point is (3 ± √5, 0)

3. (-2, 4):
x = ±√(5 - 2(4))
x = ±√(-3)
Since the square root of a negative number is not a real number, this point does not have a solution.

4. (2, 4):
x = ±√(5 - 2(4))
x = ±√(-3)
Since the square root of a negative number is not a real number, this point does not have a solution.

5. (0, 8):
x = ±√(5 - 2(8))
x = ±√(-11)
Since the square root of a negative number is not a real number, this point does not have a solution.

From the above analysis, we can conclude that the solution(s) for this system of equations is (2, ±√5), (-3 ± √5, 0).