4. Solve this equation. \( \begin{array}{l}\frac{1}{2} x-6=\frac{1}{3}(x-15) \\ \text { 5. Solve this system of equations. } \\ \left\{\begin{array}{l}3 x+4 y=36 \\ y=-\frac{1}{2} x+8\end{array}\right.\end{array} \)

To solve the equation \( \frac{1}{2} x - 6 = \frac{1}{3} (x - 15) \), first eliminate the fractions by multiplying every term by 6, the least common multiple of 2 and 3. This gives us \( 3x - 36 = 2(x - 15) \). Expanding the right side yields \( 3x - 36 = 2x - 30 \). Rearranging gives us \( 3x - 2x = 36 - 30 \), leading to \( x = 6 \). Substituting \( x \) back into the original equation confirms that \( x = 6 \) is correct. For the system of equations \( 3x + 4y = 36 \) and \( y = -\frac{1}{2}x + 8 \), we can substitute \( y \) from the second equation into the first. Replacing \( y \) gives \( 3x + 4(-\frac{1}{2}x + 8) = 36 \). Simplifying leads to \( 3x - 2x + 32 = 36 \), or \( x + 32 = 36 \). Thus, \( x = 4 \). Plugging \( x \) back into the second equation yields \( y = -\frac{1}{2}(4) + 8 = 6 \). Therefore, the solution is \( x = 4 \) and \( y = 6 \).