Which system of equations has no real solutions?(1 point)Responsesy = −x2+2 and y=x2y equals negative x squared plus 2 and y equals x squaredy = x2+1 and y =−x2 −1y equals x squared plus 1 and y equals negative x squared minus 1y= x2 and y= x +1y equals x squared and y equals x plus 1y=3x+1 and y=2x+2y equals 3 x plus 1 and y equals 2 x plus 2

Answer 1

To determine which system of equations has no real solutions, we can analyze each pair of equations:

1. y = -x² + 2 and y = x²

   • The first equation is a downward-opening parabola with its vertex at (0, 2).
   • The second equation is an upward-opening parabola with its vertex at (0, 0).
   • These two parabolas do intersect at two real points, so this system has real solutions.

2. y = x² + 1 and y = -x² - 1

   • The first equation is an upward-opening parabola with its vertex at (0, 1).
   • The second equation is a downward-opening parabola with its vertex at (0, -1).
   • The distance between the vertices and the nature of their openings (one opens up and the other opens down) shows that they are always apart; thus, this system has no real solutions.

3. y = x² and y = x + 1

   • The first equation is an upward-opening parabola and the second is a linear equation.
   • They will intersect at two real points.

4. y = 3x + 1 and y = 2x + 2

   • These are both linear equations. By rearranging, we can find their point of intersection:
     • 3x + 1 = 2x + 2 -> x = 1.
     • When x = 1, y = 4.
   • So this system has a real solution.

From the analysis above, the only pair of equations that has no real solutions is:

y = x² + 1 and y = -x² - 1.