Linear Equations Word Problems Part 2 Lesson

In our last lesson, we learned how to solve a word problem that involves a linear equation in one variable. We focused on problems such as: sums of quantities, consecutive integers, and age word problems. In this lesson, we will continue to work through applications of linear equations and focus on two challenging problem types: solving mixture word problems and solving motion word problems. Before we introduce our first problem type, let's review the six-step process for solving a word problem:

Six-step method for Applications of Linear Equations in One Variable

1. Read the problem carefully and determine what you are asked to find
   • Write down the main objective of the problem
2. Assign a variable to represent the unknown
   • If more than one unknown exists, we express the other unknowns in terms of this variable
3. Write out an equation that describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result by reading back through the problem
   • We need to make sure the answer is reasonable. In other words, if asked how many miles were driven to the store, the answer shouldn't be -300 as we can't drive a negative amount of miles.

Mixture Word Problems

A very common problem in algebra deals with mixtures. Our first task is to understand how to find the pure amount of a substance in a mixture. Let's suppose we were given the following scenario:
A chemist has 20 liters of a 35% acid solution. How much pure acid is in the solution?
What we want to find here is the amount in liters of pure acid in the solution.
We don't need any variables or fancy steps here, we are just multiplying.
35% is the amount of the solution which is pure acid. When we see the keyword "of", this tells us to multiply:
20 • 0.35 = 7
So there are 7 liters of pure acid in the solution.
This easy example demonstrates how to find the pure amount of a substance in a mixture. We can take the pure percentage as a decimal and multiply by the total amount of our mixture. Let's take a look at a harder example.
Example 1: Solve each word problem.
A chemist needs to mix 20 gallons of a 30% acid solution with some 60% acid solution to get a mixture that is 50% acid. How many gallons of the 60% acid solution should be used?
Let's think about this in terms of our six-step process.
1) The main objective here is to determine how many gallons of 60% acid solution is needed.
2) We can assign a variable like x to represent the number of gallons of 60% acid solution that is needed.
let x = # of gallons of 60% acid solution
3) We write our equation. Let's first think about the information using a table. This can sometimes help to organize our thoughts:

Gallons	Acid %	Pure Acid
20	0.30	6
x	0.60	0.6x
x + 20	0.50	0.5(x + 20)

If we look at our table, we see a column for the number of gallons, the acid % as a decimal, and the amount of pure acid in gallons. We know that the amount of pure acid in gallons from the first solution (6) added to the amount of pure acid in gallons from the second solution (0.6x) must be equal to the amount of pure acid in gallons in the final mixture [0.5(x + 20)].
This sets up our equation as:
6 + 0.6x = 0.5(x + 20) 4) Let's solve our equation:
6 + 0.6x = 0.5(x + 20)
6 + 0.6x = 0.5x + 10
6 - 6 + 0.6x = 0.5x + 10 - 6
0.6x = 0.5x + 4
0.6x - 0.5x = 0.5x - 0.5x + 4
0.1x = 4
10(0.1)x = 4(10)
x = 40
5) Since x = 40, this tells us the chemist will need 40 gallons of the 60% acid solution to create the given mixture. Let's create a nice clean sentence:
The chemist will need to use 40 gallons of the 60% acid solution.
6) Check
Let's check the % of acid in the final mixture. This brings us back to our equation:
6 + 0.6x = 0.5(x + 20)
Let's plug in a 40 for x and check:
6 + 0.6(40) = 0.5(40 + 20)
6 + 24 = 0.5(60)
30 = 30
Our solution is correct. Mixing 20 gallons of a 30% acid solution with 40 gallons of a 60% acid solution gives us a mixture that is 50% acid.

Motion Word Problems

Another common problem type is known as a motion word problem. To solve this type of problem, we need to know the formula:
d = rt
Which stands for:
distance = rate of speed x amount of time traveled
To see this in action, suppose we take a road trip and travel at an average speed of 60 mph for 7 hours. If we want the distance traveled, we can multiply 60 x 7 to get 420. This means traveling at an average speed of 60 mph for 7 hours gives us a traveled distance of 420 miles. Let's take a look at an example.
Example 2: Solve each word problem.
Beth made a trip to Jamestown and back. The trip there took three hours and the trip back took five hours. What was Beth's average speed on the trip there if she averaged 45 mph on the return trip?
Let's think about this in terms of our six-step process.
1) The main objective here is to determine Beth's average speed on the trip to Jamestown.
2) We can assign a variable like x to represent her average speed on the trip to Jamestown.
let x = the average speed of Beth in miles per hour on her trip to Jamestown
3) We write our equation. Let's first think about the information using a table. This can sometimes help to organize our thoughts:

Trip	Rate (mph)	Time (hours)	Distance (miles)
There	x	3	3x
Back	45	5	225

If we look at our table, we see a column for the rate in miles per hour, the time in hours, and the distance in miles. We know that she travels to and from the same location. Although in our everyday life we may choose a different route, we will assume that Beth travels the same route and therefore has the same distance to Jamestown and back from Jamestown. This means the distance there (3x) can be set equal to the distance coming back (225).
This sets up our equation as:
3x = 225 4) Let's solve our equation:
3x = 225
3/3 x = 225/3
x = 75
5) Since x = 75, this tells us Beth has traveled at an average speed of 75 mph on her trip there. Let's create a nice clean sentence:
Beth drove at an average speed of 75 mph on her trip to Jamestown.
6) Check
Let's check the distance to Jamestown on each leg of the trip.
Trip There: distance (225) = rate (75) x time (3)
225 = 75 x 3
225 = 225
Trip Back: distance (225) = rate (45) x time (5)
225 = 45 x 5
225 = 225
Our solution is correct. Beth drove at an average speed of 75 mph on her trip to Jamestown.

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