Houghton Mifflin Harcourt Algebra 1, 2015
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Houghton Mifflin Harcourt Algebra 1, 2015 View details
1. Creating Systems of Linear Equations
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Exercise 19 Page 439

What kind of model would you have to construct for each plane?

See solution.

Practice makes perfect

To find the time at which both planes have traveled the same distance, we would have to construct a linear model for each of the planes. Then, we would have to set and solve the system of linear equations formed by those models.

An Example

Let's look at an example. Suppose we are given a table with some information about two planes.

Distance Traveled After Takeoff Time Since Takeoff
Plane A 900 km 1.5 hours
Plane B 700 km 1 hour

Since we know the distance and time for both planes, we can find their speed. speed=distance/time Let's start by finding the speed for plane A.

speed=distance/time
speed=900/1.5
speed=600

The speed of plane A is 600 kilometers per hour. Let's do the same thing for plane B. Its distance after takeoff is 700 kilometers and the time is one hour.

speed=distance/time
speed=700/1
speed=700

The speed of plane B is 700 kilometers per hour. We can now write a system of equations. In each equation, let d be the distance traveled since takeoff in terms of the time t that the planes pass through the same point. d=600t+900 & (I) d=700t+700 & (II) Let's solve the system! Since the d-variable has a coefficient of 1 in both equations, we will use the Elimination Method. Let's subtract Equation (I) from Equation (II).

d=600t+900 d=700t+700
d=600t+900 d- d=700t+700-( 600t+900)
â–¼
(II): Solve for t
d=600t+900 d-d=700t+700-600t-900
d=600t+900 0=100t-200
d=600t+900 200=100t
d=600t+900 2=t
d=600t+900 t=2

Finally, we will substitute 2 for t in Equation (I), and solve for d.

d=600t+900 t=2
d=600( 2)+900 t=2
d=1200+900 t=2
d=2100 t=2

We found that the planes will both have traveled 2100km two hours later they pass through the same point. We can use this same process for any situation where we have the data given to us about the planes.