To determine which system of equations has no real solutions, we will analyze each pair of equations to see if their graphs intersect.
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\( y = -x^2 + 2 \) and \( y = x^2 \):
- Setting the equations equal: \[ -x^2 + 2 = x^2 \implies 2 = 2x^2 \implies x^2 = 1 \implies x = \pm 1 \]
- Both \( x = 1 \) and \( x = -1 \) provide valid intersections, giving us real solutions.
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\( y = x^2 + 1 \) and \( y = -x^2 - 1 \):
- Setting the equations equal: \[ x^2 + 1 = -x^2 - 1 \implies 2x^2 + 2 = 0 \implies x^2 = -1 \]
- No real \( x \) values satisfy this, so there are no real solutions.
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\( y = x^2 + 1 \) and \( y = -x^2 - 1 \):
- This is the same as the previous pair. It has been shown to have no real solutions.
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\( y = x^2 \) and \( y = x + 1 \):
- Setting the equations equal: \[ x^2 = x + 1 \implies x^2 - x - 1 = 0 \]
- This quadratic can be solved using the quadratic formula, leading to two real solutions.
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\( y = 3x + 1 \) and \( y = 2x + 2 \):
- Setting the equations equal: \[ 3x + 1 = 2x + 2 \implies x = 1 \]
- There is one intersection at \( x = 1 \), hence real solutions exist.
The system of equations that has no real solutions is:
\( y = x^2 + 1 \) and \( y = -x^2 - 1 \).