Harper and Audrina both track the number of miles they bike each week. Harper was already biking 40 miles per week, and started biking 20 more miles each week. Audrina was already biking 100 miles per week, and started biking 10 less miles each week. Is there a time when they bike the same number of miles each week?

The relationships between x, the number of weeks, and y, the number of miles Harper and Audrina bike each week, are graphed on the coordinate plane below. What is the point of intersection?
Harper
Audrina
0
1
2
3
4
5
6
7
8
9
10
0
10
20
30
40
50
60
70
80
90
100
Time (weeks)
Number of miles
The point of intersection is (
,
).
Nice!
Now that you have found the point of intersection, (2, 80), let’s check that work algebraically. The equations for the relationship between weeks and miles for Harper and Audrina are shown below. Since the number of miles Harper bikes will be equal to the number of miles Audrina bikes at the solution to this system, you can set the expression for the number of miles they each bike, 40+20x and 100−10x, equal to one another. Do that now.
Harper
Audrina
0
1
2
3
4
5
6
7
8
9
10
0
10
20
30
40
50
60
70
80
90
100
Time (weeks)
Number of miles

40+20x
=
100−10x
You got it!
Let’s solve the equation for x, the number of weeks when Harper and Audrina bike the same number of miles, to confirm that it’s the same as the x-coordinate of the point of intersection. Since there are x terms on both sides of the equation, let’s get rid of the x term from one of the sides. What move can you make to get rid of the x term on the right side of the equation, 10x, while keeping the equation balanced?
Harper’s miles = Audrina's miles
40+20x = 100−10x

Subtract 10x from each side

Add 10x to each side

Multiply each side by 10x

1 answer

To eliminate the \(10x\) term from the right side of the equation, the correct move is to add \(10x\) to each side of the equation. This will allow us to consolidate the \(x\) terms on one side and make it easier to solve for \(x\).

So the equation will be transformed as follows:

\[ 40 + 20x + 10x = 100 - 10x + 10x \]

This simplifies to:

\[ 40 + 30x = 100 \]

Now you can proceed to solve for \(x\).