- 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

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

- Read the problem and determine what you are asked to find
- Assign a variable to represent the unknown
- If more than one unknown exists, we express the other unknowns in terms of this variable

- Write out an equation that describes the given situation
- Solve the equation
- State the answer using a nice clear sentence
- 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.

### 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) |
---|---|---|

5 | 80% | 4 |

7 | 50% | 3.5 |

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) |
---|---|---|

12 | 12% | 1.44 |

x | 20% | .2x |

x + 12 | 14% | .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 the 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 table:

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 |

Quarter | x | .25 | .25x |

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 | |
---|---|---|---|

Bike | 0.75(x - 10) | x - 10 | 0.75 |

Bus | 0.25x | x | 0.25 |

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 ✔

#### Skills Check:

Example #1

Solve each word problem.

Jessica created a metal containing 70% iron by combining two other metals. One of these other metals weighed 1 mg and contained 86% iron. If the other weighed 4 mg, then what percent of it was iron?

Please choose the best answer.

Example #2

Solve each word problem.

A merchant ship traveled to Acolas and back. The trip there took 12 hours and the trip back took 14 hours. It averaged 2 mph faster on the trip there than on the trip back. What was the merchant ship's average speed on the outbound trip (traveling to the destination)?

Please choose the best answer.

Example #3

Solve each word problem.

A cashier at a local grocery store received $5.65 from 25 coins consisting of nickels and quarters only. How many nickels did the cashier receive?

Please choose the best answer.

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