1.1 .4 1.1 .5 Solve for \( x \) and \( y \) simultaneously \( 38-y+2=0 \) and \( y=-x^{2} + 2 x=12 \)

To solve the equations \( 38 - y + 2 = 0 \) and \( y = -x^2 + 2x = 12 \) simultaneously, we start with the first equation. Rearranging the first equation gives: \[ y = 40 \] Next, we substitute this value of \( y \) into the second equation. However, there seems to be a mistake in the presentation of the second equation as it appears to combine two expressions. It likely means \( y = -x^2 + 2x \) and subsequently equate it to 12. Let's first clarify the second expression: 1. Set \( y = -x^2 + 2x \) and then combine it with \( 40 = -x^2 + 2x \). This gives us the equation: \[ 40 = -x^2 + 2x \] Rearranging this leads to: \[ x^2 - 2x + 40 = 0 \] Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2, c = 40 \): \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 40}}{2 \cdot 1} \] \[ x = \frac{2 \pm \sqrt{4 - 160}}{2} \] \[ x = \frac{2 \pm \sqrt{-156}}{2} \] \[ x = \frac{2 \pm 2i\sqrt{39}}{2} \] \[ x = 1 \pm i\sqrt{39} \] Now, substituting these \( x \) values back into \( y = 40 \) gives us: - For \( x = 1 + i\sqrt{39} \), \( y = 40 \) - For \( x = 1 - i\sqrt{39} \), \( y = 40 \) So, our solutions are: 1. \( (1 + i\sqrt{39}, 40) \) 2. \( (1 - i\sqrt{39}, 40) \) Enjoy these complex solutions which show that mathematics can get a little wild!