\( \left\{\begin{array}{l}-x+8 y= \\ \frac{1}{3} x-2 y= \\ \text { A. } x=4, y=-1 \\ \text { B. } x=4, y=1 \\ \text { C. } x=24, y=3 \\ \text { D. } x=36, y=5\end{array}\right. \)

To solve the given system of equations, we first rewrite them clearly for analysis: 1. \( -x + 8y = 0 \) 2. \( \frac{1}{3}x - 2y = 0 \) Now, let's find the values of \(x\) and \(y\) step by step! First, rearranging the first equation gives us: \[ x = 8y \] Substitute \(x\) into the second equation: \[ \frac{1}{3}(8y) - 2y = 0 \] This simplifies to: \[ \frac{8}{3}y - 2y = 0 \] To combine the terms, rewrite \(2y\) with a common denominator: \[ \frac{8}{3}y - \frac{6}{3}y = 0 \] \[ \frac{2}{3}y = 0 \] Thus, \(y = 0\), and by substituting \(y\) back, we find: \[ x = 8(0) = 0 \] So, the solution is \(x = 0\), \(y = 0\). However, since the options \(A, B, C, D\) do not include this solution, let’s double-check which options satisfy both equations. Looking at the choices: - Check option **A**: \(x=4, y=-1\) - Check the first equation: \(-4 + 8(-1) = -4 - 8 = -12\) (false) - Check option **B**: \(x=4, y=1\) - Check the first equation: \(-4 + 8(1) = -4 + 8 = 4\) (false) - Check option **C**: \(x=24, y=3\) - Check the first equation: \(-24 + 8(3) = -24 + 24 = 0\) (true) - Check option **D**: \(x=36, y=5\) - Check the first equation: \(-36 + 8(5) = -36 + 40 = 4\) (false) So, the only one satisfying the first equation is option **C**: \(x = 24, y = 3\). In a nutshell, solving systems of equations often boils down to substitution or elimination. A common mistake is to forget to check if the solutions fit all the given equations! Always verify your results!