Which system of equations has no real solutions?(1 point)

Question

Responses

y=3x+1 and y=2x+2
y equals 3 x plus 1 and y equals 2 x plus 2

y = −x2+2 and y=x2
y equals negative x squared plus 2 and y equals x squared

y = x2+1 and y =−x2 −1
y equals x squared plus 1 and y equals negative x squared minus 1

y= x2 and y= x +1
y equals x squared and y equals x plus 1

Answer 1

To determine which system of equations has no real solutions, we need to analyze the equations given.

1. y = 3x + 1 and y = 2x + 2: These are linear equations with different slopes (3 and 2), so they will intersect at one point. Thus, they have one real solution.

2. y = -x² + 2 and y = x²: The first equation is a downward-opening parabola and the second one is an upward-opening parabola. They can intersect at two points, one point, or potentially none. We would need to solve for their intersection to determine if they do have real solutions, but generally, two parabolas can intersect in real solutions.

3. y = x² + 1 and y = -x² - 1: The first equation represents a parabola that opens upward and has its vertex at (0, 1). The second equation represents a parabola that opens downward and is shifted down (its vertex is at (0, -1)). Since the minimum point of the first is at (0, 1) and the maximum of the second is at (0, -1), these two curves do not intersect; thus, there are no points where they have the same y-value. So this system has no real solutions.

4. y = x² and y = x + 1: The first equation represents a parabola opening upward and the second equation is a line. They will intersect at some points (either zero, one, or two points), specifically since a line can cross a parabola.

Based on this analysis, the system of equations which has no real solutions is:

y = x² + 1 and y = -x² - 1.