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