Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
1. Solving Systems by Graphing
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Exercise 42 Page 369

Practice makes perfect
a

Let's call the distance from the start of the trail d and the number of hours spent hiking h.

First Hiker

The first hiker walks at a speed of 4mi/h. This means that, after h hours, he will have covered 4h miles of the trail. Furthermore, since he started 6 miles from the beginning of the trail, we can add this to 4h to get his total number of miles away from the start of the trail. d= 4h+ 6

Second Hiker

We can write an equation for the second hiker in a similar fashion. He walks at a speed of 3mi/h and starts 1 mile away from the beginning. d= 3h+ 1 We can combine our two equations to create a system of equations. d=4h+6 d=3h+1

b

Let's start by drawing the graph of d=4h+6. To do so, we will plot the y-intercept 6 and use the slope 4 to find another point on the line. Then we will connect these two points using a straightedge.

We will follow the same procedure to draw the graph of d=3h+1.

The intersection point is (- 5,- 14).

c

Because neither distance nor hours can be negative, this point of intersection means nothing in the context of the exercise. There is no meaningful solution to this system of equations.

Alternative Solution

A Logical Approach
We could have reasoned that there would be no meaningful solution to the system with basic logic. The hiker who walks at a faster pace also started farther away from the start than the slower hiker. It makes sense that the two hikers would never be at the same distance from the start of the trail.