\( x=\frac{1}{2} x+5 \) the slope \( y \)-inte.

To find the slope and the \( y \)-intercept, first, we need to rearrange the equation into slope-intercept form \( y = mx + b \). Start by isolating \( x \) on one side: \[ x - \frac{1}{2}x = 5 \quad \Rightarrow \quad \frac{1}{2}x = 5 \quad \Rightarrow \quad x = 10. \] Here, we see that \( x \) is a constant, which means there isn’t a slope as traditionally defined in linear equations because this represents a vertical line. The equation tells us that at any point on this line, \( x \) is always 10, so the \( y \)-intercept can be any point along that vertical line as \( y \) can take any value while \( x \) remains 10. In this case, you can visualize the line in a coordinate plane where the vertical line intersects the x-axis at \( x = 10 \). Therefore, the slope is undefined (as vertical lines do not have slopes), and specific \( y \)-intercepts are in the form (10, y), where y can be any real number!