1. **State the problem:** We are given the system of equations: $$x + 13 + y = 0$$ $$x + 3y + 2 = 0$$ $$x = 1$$ We need to find the area bounded by these lines. 2. **Rewrite the equations in slope-intercept form:** From the first equation: $$x + y + 13 = 0 \implies y = -x - 13$$ From the second equation: $$x + 3y + 2 = 0 \implies 3y = -x - 2 \implies y = -\frac{1}{3}x - \frac{2}{3}$$ The third equation is a vertical line: $$x = 1$$ 3. **Find the intersection points:** - Intersection of line 1 and line 2: Set their $y$ equal: $$-x - 13 = -\frac{1}{3}x - \frac{2}{3}$$ Multiply both sides by 3 to clear denominators: $$-3x - 39 = -x - 2$$ Bring all terms to one side: $$-3x + x = -2 + 39 \implies -2x = 37 \implies x = -\frac{37}{2} = -18.5$$ Find $y$: $$y = -x - 13 = -(-18.5) - 13 = 18.5 - 13 = 5.5$$ So intersection point $A = (-18.5, 5.5)$. - Intersection of line 1 and $x=1$: Substitute $x=1$ into line 1: $$y = -1 - 13 = -14$$ Point $B = (1, -14)$. - Intersection of line 2 and $x=1$: Substitute $x=1$ into line 2: $$y = -\frac{1}{3}(1) - \frac{2}{3} = -\frac{1}{3} - \frac{2}{3} = -1$$ Point $C = (1, -1)$. 4. **Calculate the area of the triangle formed by points $A$, $B$, and $C$:** Use the formula for the area of a triangle given coordinates: $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Let: $$A = (x_1, y_1) = (-18.5, 5.5)$$ $$B = (x_2, y_2) = (1, -14)$$ $$C = (x_3, y_3) = (1, -1)$$ Calculate: $$x_1(y_2 - y_3) = -18.5(-14 - (-1)) = -18.5(-13) = 240.5$$ $$x_2(y_3 - y_1) = 1(-1 - 5.5) = 1(-6.5) = -6.5$$ $$x_3(y_1 - y_2) = 1(5.5 - (-14)) = 1(19.5) = 19.5$$ Sum: $$240.5 - 6.5 + 19.5 = 253.5$$ Area: $$\frac{1}{2} |253.5| = 126.75$$ 5. **Final answer:** The area bounded by the lines is **126.75** square units.