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

Question

Responses

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

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

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

Answer 1

To determine which system of equations has no real solutions, we can look at each pair of equations and analyze their graphs.

1. y = x² and y = x + 1: The first equation is a parabola opening upwards, while the second is a line with a positive slope. They intersect at two points, so they have real solutions.

2. y = x² + 1 and y = -x² - 1: The first equation is a parabola opening upwards, and the second is a downward opening parabola. The lowest point (vertex) of y = x² + 1 is at (0, 1), while the highest point of y = -x² - 1 is at (0, -1). Since the first parabola (y = x² + 1) is always above the line y = -1 and they do not intersect, this system has no real solutions.

3. y = 3x + 1 and y = 2x + 2: Both of these are linear equations. They have different slopes (3 and 2), so they intersect at one point and thus have real solutions.

4. y = -x² + 2 and y = x²: The first equation is a downward-opening parabola, while the second is an upward-opening parabola. They intersect at two points (when both equal to each other). Therefore, this system has real solutions.

Based on the analysis, the system that has no real solutions is:

y = x² + 1 and y = -x² - 1.