1. **State the problem:** We need to find the equation of a line with slope $\frac{3}{5}$ that passes through the point $(11,6)$ in slope-intercept form $y=mx+b$. 2. **Recall the slope-intercept form:** The equation of a line is $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept. 3. **Use the point-slope form to find $b$:** Substitute $m=\frac{3}{5}$, $x=11$, and $y=6$ into $y=mx+b$: $$6=\frac{3}{5}\times 11 + b$$ 4. **Calculate $\frac{3}{5} \times 11$:** $$\frac{3}{5} \times 11 = \frac{33}{5}$$ 5. **Solve for $b$:** $$6 = \frac{33}{5} + b$$ Subtract $\frac{33}{5}$ from both sides: $$b = 6 - \frac{33}{5} = \frac{30}{5} - \frac{33}{5} = -\frac{3}{5}$$ 6. **Write the final equation:** $$y = \frac{3}{5}x - \frac{3}{5}$$ This is the equation of the line in slope-intercept form with slope $\frac{3}{5}$ passing through $(11,6)$.