### Free Educational Resources

3 Tutorials that teach Distance, Rate, and Time in a System of Equations

# Distance, Rate, and Time in a System of Equations

##### Rating:
(0)
Author: Sophia Tutorial
##### Description:

This lesson will introduce distance rate and time with systems of equations.

(more)

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*

No credit card required

28 Sophia partners guarantee credit transfer.

281 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 25 of Sophia’s online courses. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

• Distance, Rate, and Time Relationship
• Combined Rates
• Solving a Distance, Rate, and Time Problem using a System of Equations

Distance, Rate, and Time Relationship

A speed limit sign, something we see everyday, provides great insight into the relationship between distance, rate, and time.  Take for example a speed limit of 40 miles per hour.  Speed is a rate: in this case, it's the ratio between miles and hours.  Miles is a unit of distance, and hours is a unit of time.  Therefore, we can say that rate equals distance divided by time.

We can use this relationship to write a few equivalent equations:

In this lesson, we will work primarily with distance = rate • time or

Combined Rates

In some situations, more than one rate has an effect on the situation, which in turn affects how we represent the situation mathematically.  For example, imaging swimming across a river or lake.  In one direction, the current of the water might work in your favor, making you faster, or at least making it easier for you to keep a certain pace.  On the other hand, the current of the water might be working against you, making you slower, or making it harder for you to keep a certain pace.

When two rates work with each other (such as swimming with the current), we can add the two rates together.  When rates work against each other (such as swimming against the current), the opposing rate is negative, leading to an expression involving subtraction.

Solving a Distance, Rate, and Time Problem using a System of Equations

Consider the following scenario:

A swimmer can swim a half-mile with the current in 6 minutes.  However, traveling back against the current, the same distance takes 15 minutes.  Assuming the swimmer travels at a constant speed during both trips, what is the speed of the swimmer, and what is the speed of the current?

To solve this problem, we will set up two equations to be part of our system.  Both equations will be forms of d = r•t, and we'll use combined rates for the variable r.  A few things are given to us in our equation: the distance in both equations is 0.5 miles.  For the equation with rates working together, the time is 6 minutes, and in the equation with rates working against each other, the time is 15 minutes.

Before we set up our equations, it is important that we have uniform units of measure.  Because we typically express speed in miles per hour, we want to convert 6 minutes and 15 minutes into hours.  To do this, we will divide each number by 60 (because there are 60 minutes in 1 hour).  6 minutes becomes 0.1 hours, and 15 minutes becomes 0.25 hours.

Using d = rt, we can develop two equations in our system.  One equation represents swimming with the current, and one equation represents swimming against the current.

It is also important that we establish our variables:

= rate of the swimmer

= rate of the current

Now that we have a system set up, our job is to solve for the rate of the swimmer and the rate of the current.  To do this, we can create equivalent equations by dividing each equation by its respective time.  Our simplified system is now:

To solve for our rates, we notice that if we add the two equations, one of the terms will cancel, because it is positive in one equation, and negative in the other.  When we add the equations, we can solve for one of the rates:

Now that we know the value of one rate, we can substitute that value into either of our two original equations in the system to find the value of the other rate: