1. **State the problem:** Find the equation of a straight line with gradient $m=\frac{2}{3}$ passing through the point $(3,2)$ and express it in the form $y=mx+c$ and then in the form $ax+by+c=0$. 2. **Formula used:** The equation of a line with gradient $m$ passing through point $(x_1,y_1)$ is given by: $$y - y_1 = m(x - x_1)$$ 3. **Substitute values:** Here, $m=\frac{2}{3}$, $x_1=3$, and $y_1=2$. $$y - 2 = \frac{2}{3}(x - 3)$$ 4. **Simplify the right side:** $$y - 2 = \frac{2}{3}x - \frac{2}{3} \times 3$$ $$y - 2 = \frac{2}{3}x - 2$$ 5. **Add 2 to both sides:** $$y - 2 + 2 = \frac{2}{3}x - 2 + 2$$ $$y = \frac{2}{3}x$$ 6. **Equation in slope-intercept form:** $$y = \frac{2}{3}x + 0$$ 7. **Convert to standard form $ax + by + c = 0$:** Multiply both sides by 3 to clear the fraction: $$3y = 2x$$ Intermediate step with cancellation: $$\cancel{3}y = \cancel{2}x$$ Rearranged: $$2x - 3y = 0$$ **Final answers:** - Slope-intercept form: $y = \frac{2}{3}x$ - Standard form: $2x - 3y = 0$