Solve for \( x \) and \( y \) in each of the following sets of equation \( \begin{array}{ll}\text { (a) } y-2 x=2 & \text { and } x^{2}-2 x=3-y\end{array} \)

To solve the equations \( y - 2x = 2 \) and \( x^2 - 2x = 3 - y \), let's start with the first equation. From \( y - 2x = 2 \), we can express \( y \) in terms of \( x \): \[ y = 2x + 2. \] Now, we substitute \( y \) into the second equation \( x^2 - 2x = 3 - y \): \[ x^2 - 2x = 3 - (2x + 2). \] Simplifying the right side: \[ x^2 - 2x = 3 - 2x - 2, \] \[ x^2 - 2x = 1 - 2x. \] Next, we can add \( 2x \) to both sides to get rid of \( -2x \): \[ x^2 = 1. \] Now, we can solve for \( x \): \[ x^2 - 1 = 0, \] \[ (x - 1)(x + 1) = 0. \] Thus, \( x = 1 \) or \( x = -1 \). Now, let's find the corresponding \( y \) values for each \( x \): 1. For \( x = 1 \): \[ y = 2(1) + 2 = 4. \] 2. For \( x = -1 \): \[ y = 2(-1) + 2 = 0. \] So, the solutions are \( (x, y) = (1, 4) \) and \( (x, y) = (-1, 0) \).