Solve the system using any method. Explain your choice of method. 7. y=-x^2+4 y=-4x+8 x^2+3x+y=0 2x+y=5 9. 2x^2+4x-y=-2 x^2+y=2

Let's solve each system of equations step by step.<br /><br />### System 7:<br />1. \( y = -x^2 + 4 \)<br />2. \( y = -4x + 8 \)<br /><br />To solve this system, we can use substitution since both equations are already solved for \( y \).<br /><br />**Step 1: Set the two expressions for \( y \) equal to each other:**<br />\[ -x^2 + 4 = -4x + 8 \]<br /><br />**Step 2: Rearrange the equation to form a quadratic equation:**<br />\[ -x^2 + 4 = -4x + 8 \]<br />\[ -x^2 + 4x + 4 = 8 \]<br />\[ -x^2 + 4x - 4 = 0 \]<br />\[ x^2 - 4x + 4 = 0 \]<br /><br />**Step 3: Factor the quadratic equation:**<br />\[ (x - 2)^2 = 0 \]<br /><br />**Step 4: Solve for \( x \):**<br />\[ x - 2 = 0 \]<br />\[ x = 2 \]<br /><br />**Step 5: Substitute \( x = 2 \) back into one of the original equations to find \( y \):**<br />\[ y = -x^2 + 4 \]<br />\[ y = -(2)^2 + 4 \]<br />\[ y = -4 + 4 \]<br />\[ y = 0 \]<br /><br />So, the solution to the system is:<br />\[ (x, y) = (2, 0) \]<br /><br />### System 9:<br />1. \( 2x^2 + 4x - y = -2 \)<br />2. \( x^2 + y = 2 \)<br /><br />To solve this system, we can use substitution or elimination. Here, we'll use substitution.<br /><br />**Step 1: Solve the second equation for \( y \):**<br />\[ y = 2 - x^2 \]<br /><br />**Step 2: Substitute \( y = 2 - x^2 \) into the first equation:**<br />\[ 2x^2 + 4x - (2 - x^2) = -2 \]<br /><br />**Step 3: Simplify the equation:**<br />\[ 2x^2 + 4x - 2 + x^2 = -2 \]<br />\[ 3x^2 + 4x - 2 = -2 \]<br /><br />**Step 4: Rearrange the equation to form a quadratic equation:**<br />\[ 3x^2 + 4x - 2 = -2 \]<br />\[ 3x^2 + 4x = 0 \]<br /><br />**Step 5: Factor the quadratic equation:**<br />\[ x(3x + 4) = 0 \]<br /><br />**Step 6: Solve for \( x \):**<br />\[ x = 0 \quad \text{or} \quad 3x + 4 = 0 \]<br />\[ x = 0 \quad \text{or} \quad x = -\frac{4}{3} \]<br /><br />**Step 7: Substitute \( x = 0 \) and \( x = -\frac{4}{3} \) back into \( y = 2 - x^2 \) to find \( y \):**<br /><br />For \( x = 0 \):<br />\[ y = 2 - (0)^2 \]<br />\[ y = 2 \]<br /><br />For \( x = -\frac{4}{3} \):<br />\[ y = 2 - \left(-\frac{4}{3}\right)^2 \]<br />\[ y = 2 - \frac{16}{9} \]<br />\[ y = \frac{18}{9} - \frac{16}{9} \]<br />\[ y = \frac{2}{9} \]<br /><br />So, the solutions to the system are:<br />\[ (x, y) = (0, 2) \quad \text{and} \quad \left(-\frac{4}{3}, \frac{2}{9}\right) \]<br /><br />### Summary:<br />- For System 7: The solution is \( (x, y) = (2, 0) \).<br />- For System 9: The solutions are \( (x, y) = (0, 2) \) and \( \left(-\frac{4}{3}, \frac{2}{9}\right) \).