Lesson Objectives
• Demonstrate an understanding of how to translate phrases into algebraic expressions and equations
• Demonstrate an understanding of the six-step method used for solving applications of linear equations
• Learn how to solve mixture word problems
• Learn how to solve coin word problems
• Learn how to solve motion word problems

## How to Solve Word Problems with Linear Equations

In our last lesson, we reviewed the six-step method used to solve a word problem that involves a linear equation in one variable.

### Six-step method for Solving Word Problems with Linear Equations in One Variable

1. Read the problem and determine what you are asked to find
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 which describes the given situation
4. Solve the equation
5. State the answer using a nice clear sentence
6. Check the result
• We need to make sure the answer is reasonable. In other words, if asked how many students were on a bus, the answer shouldn't be (-4) as we can't have a negative amount of students on a bus.
In this lesson, we will look at mixture word problems, coin word problems, and motion word problems. Let's begin by learning how to solve a mixture word problem.

### Mixture Word Problems

Mixture word problems can be very frustrating. The key is to understand how to find the concentration of a particular substance within a given mixture. Let's start out with a simple example.
Example 1: Solve each word problem
An acid solution was made by mixing 5 gallons of an 80% acid solution and 7 gallons of a 50% acid solution. Find the concentration of acid in the new mixture.
For this problem, we don't need any variables. Let's think for a moment about how much acid is in each mixture:
Mixture (Gal) Acid % Pure Acid (Gal)
580%4
750%3.5
If we look at our table above, we can see that there are 4 gallons of pure acid in our 5-gallon mixture and 3.5 gallons of pure acid in our 7-gallon mixture. If we combine the two mixtures, we get a 12-gallon mixture which contains 7.5 gallons of pure acid. How can we find the % of acid for our new mixture? We divide the number of gallons of pure acid by the number of gallons of the total mixture:
7.5 ÷ 12 = 0.625
We can say our acid percentage is 62.5% for the mixture. The main thing is to understand that we can take the pure amount of a substance and divide it by the total mixture to find a concentration. Let's look at a more challenging mixture word problem.
Example 2: Solve each word problem
A local chemist made a solution that was 14% alcohol. He started out with 12 gallons of a 12% alcohol solution. He then added an unknown number of gallons of a 20% acid solution. How many gallons of the 20% acid solution did the chemist add to the initial alcohol solution?
Step 1) After reading the problem, it is clear that we want to find the number of gallons of the 20% acid solution that needs to be added to the 12% acid solution in order to obtain a solution that is 14% alcohol.
Step 2) let x = number of gallons of the 20% alcohol solution
Step 3) To write out an equation, let's first organize our information in a table:
Mixture (Gal) Alcohol % Pure Alcohol (Gal)
1212%1.44
x20%.2x
x + 1214%.14(x + 12)
From our table, we can see the number of gallons of pure acid from the 12% acid solution is 1.44 (12 • 0.12 = 1.44). Additionally, we represent the number of gallons of pure acid from our 20% acid solution as 0.2x. These two amounts must be equal to the amount of pure acid in the final solution, which is .14(x + 12).
(.14 » 14% acid concentration in the final solution)
(x + 12) » the number of gallons in the final solution
Let's set up our equation:
.14(x + 12) = .2x + 1.44
Step 4) Solve the equation
.14(x + 12) = .2x + 1.44
.14x + 1.68 = .2x + 1.44
Multiply each side by 100 to clear the decimals:
14x + 168 = 20x + 144
14x - 20x = 144 - 168
-6x = -24
x = -24 ÷ -6
x = 4
Step 5) Since x is the number of gallons of the 20% alcohol solution used, we can state our answer as:
The chemist used 4 gallons of the 20% alcohol solution.
Step 6) We can check our result by reading through the problem. We know the end result is a 16-gallon solution that is 14% alcohol. We can compare this to the sum of the alcohol from the two different solutions:
Solution 1 (12 gallons, 12% alcohol):
12 • .12 = 1.44
Solution 2 (4 gallons, 20% alcohol):
4 • .2 = .8
1.44 + 0.8 = 2.24
There are 16 gallons in the final solution with an alcohol percentage of .14, let's check to see if the number of gallons matches up:
16 • 0.14 = 2.24
2.24 = 2.24

### Solving Coin Word Problems

Another very common word problem is known as a coin word problem. We may also see this same type of problem occur with paper bills as well. Essentially this type of problem will give us a value for all of the coins or bills in the problem. Our goal will be to find the individual number of coins or bills involved. Let's look at an example.
Example 3: Solve each word problem
At the end of Jessica's shift, she counted down her register. She had a total of $14.25 in change (coins). This change consisted of nickels, dimes, and quarters only. There were three times as many quarters as dimes and two-thirds the number of nickels as quarters. How many of each type of coin did Jessica have in her register? Step 1) After reading the problem, it is clear that we want to find the number of nickels, dimes, and quarters that were in Jessica's register Step 2) When we have more than one unknown, we can let a variable represent one of the unknowns and then model the other unknowns based on our variable. In this case, quarters are used in both comparisons. let x = # of quarters that were present in Jessica's register Then (1/3)x = # of dimes that were present in Jessica's register (since the number of quarters was 3 times more than the number of dimes) Then (2/3)x = # of nickels that were present in Jessica's register (since the number of nickels was 2/3 of the number of quarters) Step 3) To make an equation, we have to think about the information given. Many people make the mistake of trying to just add the amounts we just came up with (1/3)x, (2/3)x, and x and set this equal to 14.25. This will not work as we are comparing a number to a value. In other words,$14.25 is the value of the coins, whereas the sum of (1/3)x, (2/3)x, and x would be a number of coins. In this case and with most coin or denomination of money problems, we have to take an extra step and multiply each coin by its value. Let's look at the information organized in a chart:
Coin Number of Coins Value of Each Coin Total Value
Nickel(2/3)x.05(1/30)x
Dime(1/3)x.10(1/30)x
Quarterx.25.25x
To make things easier, we can use 1/20 in the place of .05, 1/10 in the place of .1 and 1/4 in the place of .25.
Total Value:
Nickels: $$\frac{1}{20}\cdot \frac{2}{3}x=\frac{1}{30}x$$ Dimes: $$\frac{1}{10}\cdot \frac{1}{3}x=\frac{1}{30}x$$ Quarters: $$\frac{1}{4}x$$ Let's set up our equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ Step 4) Solve the equation: $$\frac{1}{30}x + \frac{1}{30}x + \frac{1}{4}x=14.25$$ $$\frac{2}{30}x + \frac{1}{4}x=14.25$$ $$\frac{1}{15}x + \frac{1}{4}x=14.25$$ Clear the fractions, multiply each side by 60: $$4x + 15x=855$$ $$19x=855$$ $$x=45$$ Step 5) Since x represented the number of quarters, this tells us Jessica had 45 quarters. She had 2/3 the number of nickels as quarters. This means she had 30 nickels (45 • 2/3). Lastly, she had 1/3 the number of dimes as quarters. This means she had 15 dimes (45 • 1/3). We can state our answer as:
Jessica had 30 nickels, 15 dimes, and 45 quarters in her register.
Step 6) We can check to see if the value matches.
Nickels: 30 • .05 = 1.5
Dimes: 15 • .1 = 1.5
Quarters: 45 • .25 = 11.25
Sum the amounts:
1.5 + 1.5 + 11.25 = 14.25
14.25 = 14.25
We can also check the number of coins: we are told there are three times as many dimes as quarters:
3 • 15 = 45
45 = 45
We are also told there are two-thirds the number of nickels as quarters:
45 • 2/3 = 30
30 = 30

### Solving Motion Word Problems » d = r x t

At this point, most of us have seen motion word problems. These problems use the distance formula (d = r x t) in order to gain a solution. The distance formula:
d » distance traveled
r » rate of speed
t » time traveled
This formula is one of the most intuitive. If we take a road trip and travel at an average rate of speed of 60 miles per hour for 12 hours, we can calculate our distance traveled as:
d = r • t
d = 60 • 12
d = 720
So in this example, we would have traveled a total distance of 720 miles. Let's look at an example.
Example 4: Solve each word problem
Mishel can get to work in 15 minutes when she takes the bus. When she rides her bike to work, it takes 45 minutes. Her average speed when riding her bike is 10 miles per hour slower than the speed of the bus. What is the average speed of the bus?
Step 1) After reading the problem, it is clear that we need to find the average speed of the bus in miles per hour
Step 2) In this case, we have quite a few things that are unknown. We are given the time it takes to get to work by bus (15 minutes) and by bike (45 minutes). We are not given a rate for either scenario. We are just told that whatever the rate is for the bus, the bike is 10 miles per hour slower. If we let a variable like x be equal to the speed of the bus, then (x - 10) would be the speed of the bike.
let x = speed of the bus in miles per hour
then: x - 10 = speed of the bike in miles per hour
It is also important to note that for this type of problem we assume the distance to work is always the same. This may not be very realistic as a bike path in the real world may not be exactly the same as the path taken for a car. Here we will just say the distance to work is always the same.
Step 3) To make an equation, let's look at our information in a table. Remember our distance formula is: d = r x t. This means we can multiply rate of speed by time traveled to obtain a distance.
d r t
Bike0.75(x - 10)x - 100.75
Bus0.25xx0.25
First and foremost, looking at the table might cause some confusion. Why is time set to 0.75 for the bike and 0.25 for the bus? We did this because we are working with a speed that is in miles per hour. Therefore, we converted our time into hours. 45 minutes is the same as 3/4 of an hour or 0.75 hours. Similarly, 15 minutes is 1/4 of an hour or 0.25 hours. Now that we have modeled our situation, we can set up an equation. We know the distances are equal for the two scenarios. Let's set those equal to each other and solve:
0.75(x - 10) = 0.25x
Step 4) Solve the equation:
0.75x - 7.5 = 0.25x
Multiply by 100 to clear the decimals:
75x - 750 = 25x
75x - 25x = 750
50x = 750
x = 15
Step 5) Since x represented the speed of the bus in miles per hour, this tells us the bus travels at an average speed of 15 miles per hour and the bike will travel at (15 - 10 = 5) 5 miles per hour. Let's state our answer as:
The bus travels at an average speed of 15 miles per hour, while Mishel's bike travels at an average speed of 5 miles per hour.
Step 6) To check our answer, let's see if our answer matches up with the information given.
The distance traveled to work should be the same in each case:
Bike:
0.75 hours • 5 miles per hour
Bus:
0.25 hours • 15 miles per hour
0.75 • 5 = 0.25 • 15
3.75 = 3.75