Which systems of equations have no real number solutions? Check all that apply. y = x2 + 4x + 7 and y = 2 y = x2 – 2 and y = x + 5 y = –x2 – 3 and y = 9 + 2x y = –3x – 6 and y = 2x2 – 7x y = x2 and y = 10 – 8x

By checking it analytically and graphically, we will see that the systems with no real solutions are the 1, 3, and 4 systems.When a system has no real solutions?A system of equations has no solutions when the equations never do intercept, we can check it graphically or analytically, I will use both methods so you can see how each one is used.1) The first system is:y = x^2 + 4x + 7y = 2The two equations only intercept if there is a value of x such that:x^2 + 4x + 7 = 2x^2 + 4x + 7 - 2 = 0x^2 + 4x+ 5 = 0Does this has real solutions? we need to check the discriminant:4^2 - 4*1*5 = 16 - 20 = -4The discriminant is negative, so this has no real solutions, then there are no real values of x such that:x^2 + 4x + 7 = 2Meaning that the system has no real solutions.2) The second system is:y = x^2 - 2y = x + 5We do the same thing than before:x^2 - 2 = x + 5x^2 - x - 5 - 2 = 0x^2 - x - 7Now we check the discriminant:(-1)^2 - 4*(-7)*1 = 1 + 28 = 29The discriminant is positive, so this system has real solutions.3) The third system is:y = -x^2 - 3y = 9 + 2xWe do the same thing:-x^2 - 3 = 9 + 2x-x^2 - 3 - 9 - 2x = 0-x^2 - 2x - 12 = 0Now let's check the discriminant:(-2)^2 - 4*(-1)*(-12) = 4 - 48 = -44The discriminant is negative, so there are no real solutions.4) y = -3x - 6    y = 2x^2 - 7xSame thing:-3x - 6 = 2x^2 - 7x-3x - 6 - 2x^2 + 7x = 0-2x^2 + 4x - 6 = 0Again, let's check the discriminant:4^2 - 4*(-2)*(-6) = 16 - 48 = -32This system has no real solutions.5) y = x^2    y = 10 - 8xNow, let's check this graphically, we just need to graph both functions and see if they intersect. The graph can be seen below, there you can see that the curves intersect, so this system has real solutions.If you want to learn more about systems of equations, you can read: