b) \[ \begin{array}{l} x+2 y=4 \\ x-2 y=0 \end{array} \] d) \[ \begin{array}{l} x+4 y=14 \\ 3 x+2 y=12 \end{array} \]

To solve the first system of equations, you can either use substitution or elimination. Here, let's use elimination. Adding the two equations: 1) \( x + 2y = 4 \) 2) \( x - 2y = 0 \) Adding them gives \( 2x = 4 \), so \( x = 2 \). Plugging \( x \) back into either equation (let's choose the first one) gives \( 2 + 2y = 4 \), leading us to \( 2y = 2 \) or \( y = 1 \). Thus, the solution is \( (2, 1) \). Now for the second system: 1) \( x + 4y = 14 \) 2) \( 3x + 2y = 12 \) Let's solve for \( x \) in the first equation: \( x = 14 - 4y \). Substitute \( x \) into the second equation: \( 3(14 - 4y) + 2y = 12 \) Distributing gives \( 42 - 12y + 2y = 12 \) or \( 42 - 10y = 12 \). Solving for \( y \) leads to \( 10y = 30 \), so \( y = 3 \). Plugging \( y \) back into \( x = 14 - 4(3) \) yields \( x = 2 \). Thus, the solution is \( (2, 3) \).