# Chapter 6 - Systems of Equations and Inequalities - 6-1 Solving Systems by Graphing - Practice and Problem-Solving Exercises - Page 369: 42

a.) $d=4t+6$ $d=3t+1$ b.) Refer to Image c.) No. As the intersection point is $(-5,-14)$ it implies that the two lines intersect when in reality, the two hikers will never actually meet, as time cannot be negative ($-5<0$). a.) For Hiker 1: $s = 4mi/h$ and $d_{0} = 6mi$ This implies, $d=4t+6$ For Hiker 2: $s=3mi/h$ and $d_{0}=1mi$ This implies, $d=3t+1$ Therefore, the system of equations becomes, $$d=4t+6$$ $$d=3t+1$$ NOTE: We are assuming that the distance from the beginning is a positive value since the trail is a "marked" one, implying that the hikers must be a positive distance from the start. b.) Use any graphing method (hand-drawn or digital) to plot $y=4x+6$ and $y=3x+1$ to find the intersecting point (use $x$ and $y$ to avoid confusion). Refer to the image. c.) The point of intersection for $d=4t+6$ and $d=3t+1$ is $(-5,-14)$ as seen in b.). 