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

Answer 1

To determine which system of equations has no real solutions, we will analyze each pair of equations to see if their graphs intersect.

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

2. \( y = x^2 + 1 \) 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.

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

   • This is the same as the previous pair. It has been shown to have no real solutions.

4. \( y = x^2 \) 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.

5. \( y = 3x + 1 \) 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 \( y = -x^2 - 1 \).