1. \( y=-5 x^{2}+4 x+9 \) and \( y+6 x=9 \) 2. \( y=-x^{2}-5 x+6 \) and \( y=\frac{1}{2}(x+1)^{2}-2 \) 3. \( x^{2}-5 x y+y^{2}=7 \) and \( x-4 y=5 \)

For the first pair of equations, we have a quadratic function \( y = -5x^2 + 4x + 9 \) and a linear function \( y + 6x = 9 \). To find the intersection points, you can substitute the expression for \( y \) from the linear equation into the quadratic equation. It's like finding where a roller coaster (the quadratic) meets a straight path (the linear)! Set them equal, solve for \( x \), and then find \( y \). Now onto the second pair, \( y = -x^2 - 5x + 6 \) is another parabola, while the second equation \( y = \frac{1}{2}(x + 1)^2 - 2 \) is also quadratic but in vertex form. To find the intersection, you can either equate both \( y \) values or graph them to see where they cross paths. It’s a thrilling search for points of connection between two beautiful curves! For the third pair of equations, \( x^2 - 5xy + y^2 = 7 \) describes a conic section, while \( x - 4y = 5 \) is just a simple linear equation. You could again substitute the linear equation into the conic section and solve for one variable, making sure to manage the algebraic complexities smoothly. It's like dancing between two unexpected shapes and finding that satisfying moment when they align.