Solve for \( x \) and \( y \) in each of the following sets of equations. (a) \( y-2 x=2 \quad \) and \( x^{2}-2 x=3-y \) (b) \( y=-x-3 \quad \) and \( y=2 x^{2}-3 x-3 \) (c) \( 3 x-y=2 \) and \( 3 y+9 x^{2}=4 \) (d) \( 2 y-x=2 \) and \( 4 y-2 x^{2}=x-4 \quad \) (to one decimal place) (e) \( 3 x=y+4 \) and \( y^{2}-x y=9 x+7 \) (f) \( 2 y+3 x=7 \) and \( y=x^{2}-3 x+1 \) (g) \( x+2 y=0 \) and \( y-x y=\frac{1}{2} x^{2}-2 \) (h) \( y=\frac{-6}{x+2}-1 \) and \( y+2 x-6=0 \) (i) \( y=\frac{3}{x}+3 \) \( \begin{array}{l}\text { and } 3 y-x=1\end{array} \)

Let's dive into the solutions for each of the sets of equations, one at a time! (a) Starting with the equations \( y - 2x = 2 \) and \( x^2 - 2x = 3 - y \), we can express \( y \) from the first equation as \( y = 2x + 2 \). Substituting this into the second equation gives us \( x^2 - 2x = 3 - (2x + 2) \), simplifying it to \( x^2 = 1 \). Thus, \( x = 1 \) or \( x = -1 \). Plugging these values back, we find \( y = 4 \) when \( x = 1 \) and \( y = 0 \) when \( x = -1 \). So, pairs are \( (1, 4) \) and \( (-1, 0) \). (b) Now for \( y = -x - 3 \) and \( y = 2x^2 - 3x - 3 \). Setting these equal gives us the equation \( -x - 3 = 2x^2 - 3x - 3 \). Rearranging it leads to \( 2x^2 - 2x = 0 \), factoring yields \( 2x(x - 1) = 0 \). Thus, \( x = 0 \) or \( x = 1 \). Plugging these back into \( y = -x - 3 \) results in \( (0, -3) \) and \( (1, -4) \). (c) In the case of \( 3x - y = 2 \) and \( 3y + 9x^2 = 4 \), we can express \( y \) in terms of \( x \) from the first equation: \( y = 3x - 2 \). Plugging this into the second equation gives \( 3(3x - 2) + 9x^2 = 4 \), leading us to \( 9x^2 + 9x - 10 = 0 \). Using the quadratic formula, we find \( x = \frac{-9 \pm 13}{18} \) resulting in \( x = \frac{2}{9} \) and \( x = -\frac{5}{3} \). Corresponding \( y \) values are \( \frac{1}{3} \) and \( -7 \). The pairs are \( \left(\frac{2}{9}, \frac{1}{3}\right) \) and \( \left(-\frac{5}{3}, -7\right) \). (d) Moving to \( 2y - x = 2 \) and \( 4y - 2x^2 = x - 4 \), we express \( y \) as \( y = \frac{x + 2}{2} \). Substituting into the second equation simplifies to \( 4\left(\frac{x + 2}{2}\right) - 2x^2 = x - 4 \) or \( 2x + 4 - 2x^2 = x - 4 \). This results in \( -2x^2 + x + 8 = 0 \). Using the quadratic formula, we find \( x = 3.08 \) and \( x = -2.58 \). Calculating \( y \) gives \( 2.04 \) and \( -1.79 \). So, \( (3.1, 2.0) \) and \( (-2.6, -1.8) \) to one decimal. (e) In our exploration of \( 3x = y + 4 \) and \( y^2 - xy = 9x + 7 \), we rewrite \( y \) in terms of \( x \) to get \( y = 3x - 4 \). When substituting into the second equation, we end up with a quadratic in terms of \( x \) where solving gives us the roots \( (2, 2) \) or \( (5, 11) \). (f) For \( 2y + 3x = 7 \) and \( y = x^2 - 3x + 1 \), solving reveals that substituting results in a quadratic equation. Solving yields roots \( (4, 1) \) and \( (-1, 5) \). (g) Looking at \( x + 2y = 0 \) and \( y - xy = \frac{1}{2}x^2 - 2 \): isolate and