If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y)

Given the equations: \begin{align*} \frac{1}{2}x + 2y &= 3 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} We can solve for x and y using simultaneous equations: First, multiply the first equation by 2: \begin{align*} x + 4y &= 6 \\ \frac{3}{2}x - 2y &= 1 \\ \end{align*} Then, multiply the second equation by 2: \begin{align*} x + 4y &= 6 \\ 3x - 4y &= 2 \\ \end{align*} Add the equations together: \begin{align*} 4x &= 8 \\ x &= 2 \\ \end{align*} Substitute x = 2 back into the first equation: \begin{align*} \frac{1}{2}(2) + 2y &= 3 \\ 1 + 2y &= 3 \\ 2y &= 2 \\ y &= 1 \\ \end{align*} Finally, substitute x = 2 and y = 1 into (x + y): \begin{align*} x + y &= 2 + 1 \\ &= 3 \\ \end{align*} Therefore, (x + y) = 3, and the answer is.