1. **State the problem:** Find the area of quadrilateral O C B E formed by the intersection of lines $l_1: y=3x+5$ and $l_2: y=-2x+6$ and the coordinate axes. 2. **Find points of intersection with axes:** - For $l_1$, x-intercept $D$ is where $y=0$: $$0=3x+5 \Rightarrow x=-\frac{5}{3}$$ - For $l_1$, y-intercept $E$ is where $x=0$: $$y=3(0)+5=5$$ - For $l_2$, x-intercept $C$ is where $y=0$: $$0=-2x+6 \Rightarrow x=3$$ - For $l_2$, y-intercept $A$ is where $x=0$: $$y=-2(0)+6=6$$ 3. **Find intersection point $B$ of $l_1$ and $l_2$:** Set $3x+5 = -2x+6$: $$3x+5 = -2x+6$$ $$3x + 2x = 6 - 5$$ $$5x = 1$$ $$x=\frac{1}{5}$$ Substitute $x=\frac{1}{5}$ into $l_1$: $$y=3\times \frac{1}{5} + 5 = \frac{3}{5} + 5 = \frac{3}{5} + \frac{25}{5} = \frac{28}{5}$$ So, $B=\left(\frac{1}{5}, \frac{28}{5}\right)$. 4. **Identify points of quadrilateral $O C B E$:** - $O=(0,0)$ - $C=(3,0)$ - $B=\left(\frac{1}{5}, \frac{28}{5}\right)$ - $E=(0,5)$ 5. **Calculate area using Shoelace formula:** $$\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 - (y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1) \right|$$ Label points in order: $O(0,0), C(3,0), B\left(\frac{1}{5}, \frac{28}{5}\right), E(0,5)$ Calculate: $$x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1 = 0\times0 + 3\times \frac{28}{5} + \frac{1}{5} \times 5 + 0 \times 0 = 0 + \frac{84}{5} + 1 + 0 = \frac{84}{5} + 1 = \frac{84}{5} + \frac{5}{5} = \frac{89}{5}$$ $$y_1x_2 + y_2x_3 + y_3x_4 + y_4x_1 = 0\times3 + 0 \times \frac{1}{5} + \frac{28}{5} \times 0 + 5 \times 0 = 0$$ Area: $$\frac{1}{2} \left| \frac{89}{5} - 0 \right| = \frac{1}{2} \times \frac{89}{5} = \frac{89}{10} = 8.9$$ **Final answer:** The area of quadrilateral $O C B E$ is $\boxed{\frac{89}{10}}$ or $8.9$ square units.