Steven and Irene both track the number of miles they run each week. Steven was already running 20 miles per week, and started running 4 more miles each week. Irene was already running 30 miles per week, and started running 2 more miles each week. Is there a time when they run the same number of miles each week?

Question

The relationships between x, the number of weeks, and y, the number of miles Steven and Irene run each week, are graphed on the coordinate plane below. What is the point of intersection?
Steven
Irene
0
1
2
3
4
5
6
7
8
9
10
0
8
16
24
32
40
48
56
64
72
80
Time (weeks)
Number of miles
The point of intersection is (
,
).
Great job!
Now that you have found the point of intersection, (5, 40), let’s check that work algebraically. The equations for the relationship between weeks and miles for Steven and Irene are shown below. Since the number of miles Steven runs will be equal to the number of miles Irene runs at the solution to this system, you can set the expression for the number of miles they each run, 20+4x and 30+2x, equal to one another. Do that now.
Steven
Irene
0
1
2
3
4
5
6
7
8
9
10
0
8
16
24
32
40
48
56
64
72
80
Time (weeks)
Number of miles

20+4x =
30+2x
Excellent!
Let’s solve the equation for x, the number of weeks when Steven and Irene run 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, 2x, while keeping the equation balanced?
Steven’s miles = Irene's miles
20+4x = 30+2x

Multiply each side by 2x

Subtract 2x from each side

Add 2x to each side
You got it!
We can subtract 2x on both sides to get rid of the 2x on the right while keeping the equation balanced. Record subtracting 2x on both sides of the equation.
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−

2x
−

2x
Excellent!
Subtract 2x on both sides and rewrite the equation to represent the remaining values.
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−
2x
−
2x

=
Nice!
We have all of the variables on the left. Now, keep going. What move can you make next to get rid of 20 on the left so that the x term, 2x, is alone on the left?
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−
2x
−
2x

20+2x
=
30

Add 2x on both sides.

Subtract 20 from both sides

Add 20 to both sides
Excellent!
You can subtract 20 on both sides so that the x term remains alone on the left, and the equation is balanced. Record that in the equation.
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−
2x
−
2x

20 + 2x
=
30
−

20
−

20
Nice!
Subtract 20 from both sides and rewrite the equation to represent the remaining values.
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−
2x
−
2x

20 + 2x
=
30
−
20
−
20

=
You got it!
Keep going. Make a balanced move so that x is alone on the left, and then rewrite the equation to solve for the value of x.
Steven's miles
=
Irene's miles

20+4x
=
30+2x
−
2x
−
2x

20 + 2x
=
30
−
20
−
20

2x
=
10

x
=
Good work!
You just found that x = 5. That’s the same as the x-coordinate at the point of intersection. Let’s see if the corresponding y-coordinate is the same, too. What’s the value of y when x = 5? You can use either equation to find out. You can solve on paper, and then enter your answer on Zearn.
Miles Steven runs per week: y= 20+4x
Miles Irene runs per week: y= 30+2x

y= 20+4x
y= 30+2x
When x= 5, y=

Answer 1

To find the value of \( y \) when \( x = 5 \), you can use either equation.

Let’s use Steven's equation:

\[ y = 20 + 4x \] Substituting \( x = 5 \):

\[ y = 20 + 4(5) \] \[ y = 20 + 20 \] \[ y = 40 \]

Now, let’s check it using Irene's equation:

\[ y = 30 + 2x \] Substituting \( x = 5 \):

\[ y = 30 + 2(5) \] \[ y = 30 + 10 \] \[ y = 40 \]

In both cases, when \( x = 5 \), \( y = 40 \).

So, the corresponding value of \( y \) is 40.

The final coordinates of the point of intersection are:

\[ (5, 40) \]