Question

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

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

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

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

y = x2+1
and y =−x2 −1

Answer 1

To determine which system of equations has no real solutions, we need to analyze each pair to see if they intersect at any points.

1. **\( y = -x^2 + 2 \) and \( y = x^2 \)**:
- Setting them equal:
\[-x^2 + 2 = x^2\]
\[2 = 2x^2\]
\[x^2 = 1\]
- This gives solutions \( x = 1 \) and \( x = -1 \), so there are real solutions.

2. **\( y = x^2 \) and \( y = x + 1 \)**:
- Setting them equal:
\[x^2 = x + 1\]
\[x^2 - x - 1 = 0\]
- The discriminant \( b^2 - 4ac = (-1)^2 - 4(1)(-1) = 1 + 4 = 5 \) is positive, which means there are two real solutions.

3. **\( y = 3x + 1 \) and \( y = 2x + 2 \)**:
- Setting them equal:
\[3x + 1 = 2x + 2\]
\[3x - 2x = 2 - 1\]
\[x = 1\]
- This gives a real solution \( x = 1 \).

4. **\( y = x^2 + 1 \) and \( y = -x^2 - 1 \)**:
- Setting them equal:
\[x^2 + 1 = -x^2 - 1\]
\[x^2 + x^2 = -1 - 1\]
\[2x^2 = -2\]
\[x^2 = -1\]
- Since \( x^2 = -1 \) has no real solutions (the solutions are complex), this system has no real solutions.

Therefore, the system of equations with no real solutions is:

**\( y = x^2 + 1 \) and \( y = -x^2 - 1 \)**.