91Ó°ÊÓ

First, let's write down the given equations: 1) \( y = -x^2 + 7x - 5 \) 2) \( y = 2x + 2 \)

To apply the substitution method, substitute the expression for y from equation (2) into equation (1): \( -x^2 + 7x - 5 = 2x + 2 \)

Now, we need to re-arrange the equation to get a quadratic equation in the form of \( Ax^2 + Bx + C = 0 \) and then solve for x: \( -x^2 + 7x - 5 - 2x - 2 = 0 \) Which simplifies to: \( -x^2 + 5x - 7 = 0 \) Now, solve the quadratic equation. In this case, it's not factorizable, so we will have to use the quadratic formula: \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) Where \( A=-1, B=5, C=-7 \). So: \( x = \frac{-5 \pm \sqrt{5^2 - 4(-1)(-7)}}{-2} \) \( x = \frac{-5 \pm \sqrt{25 - 28}}{-2} \) \( x = \frac{-5 \pm \sqrt{-3}}{-2} \) Since we have a negative number inside the square root, the equation has no real solutions. Consequently, the given system of equations has no point of intersection and therefore no real solution.

The given system of equations has no real solution.