1. **State the problem:** We need to graph the two linear equations $y=3x$ and $y=-x+400$ on the same coordinate plane. 2. **Recall the form of linear equations:** Both equations are in slope-intercept form $y=mx+b$, where $m$ is the slope and $b$ is the y-intercept. 3. **Analyze the first equation $y=3x$:** - Slope $m=3$ means the line rises 3 units for every 1 unit it moves right. - Y-intercept $b=0$ means the line passes through the origin $(0,0)$. 4. **Analyze the second equation $y=-x+400$:** - Slope $m=-1$ means the line falls 1 unit for every 1 unit it moves right. - Y-intercept $b=400$ means the line crosses the y-axis at $(0,400)$. 5. **Plot key points for $y=3x$:** - At $x=0$, $y=0$. - At $x=1$, $y=3$. - At $x=2$, $y=6$. 6. **Plot key points for $y=-x+400$:** - At $x=0$, $y=400$. - At $x=100$, $y=300$. - At $x=200$, $y=200$. 7. **Draw both lines using these points.** 8. **Find the intersection point by solving the system:** Set $3x = -x + 400$. $$3x + x = 400$$ $$4x = 400$$ $$x = 100$$ Substitute $x=100$ into $y=3x$: $$y = 3 \times 100 = 300$$ So, the lines intersect at $(100, 300)$. **Final answer:** The two lines $y=3x$ and $y=-x+400$ intersect at the point $(100, 300)$ and can be graphed using their slopes and intercepts as described.