Algebra and word problems

REAL WORLD PROBLEMS:
How to Write Equations Based on Algebra Word Problems

I know that you often sit in class and wonder, «Why am I forced to learn about equations, Algebra and variables?»

But… trust me, there are real situations where you will use your
knowledge of Algebra and solving equations to solve a problem that is
not school related. And… if you can’t, you’re going to wish that you
remembered how.

It might be a time when you are trying to figure out how much you
should get paid for a job, or even more important, if you were paid
enough for a job that you’ve done. It could also be a time when you are
trying to figure out if you were over charged for a bill.

This is important stuff — when it comes time to spend YOUR money —
you are going to want to make sure that you are getting paid enough and
not spending more than you have to.

Ok… let’s put all this newly learned knowledge to work.

Click here if you need to review how to solve equations.

There are a few rules to remember when writing Algebra equations:

Let’s look at an example of an algebra word problem.

Example 1: Algebra Word Problems

Linda was selling tickets for the school play.  She sold 10 more adult tickets than children
tickets and she sold twice as many senior tickets as children tickets.

1. Let x represent the number of children’s tickets sold.
2. Write an expression to represent the number of adult tickets sold.
3. Write an expression to represent the number of senior tickets sold.
4. Adult tickets cost $5, children’s tickets cost $2, and senior tickets cost $3. Linda made $700. Write an equation to represent the total ticket sales.
5. How many children’s tickets were sold for the play? How many adult tickets were sold? How many senior tickets were sold?
   

As you can see, this problem is massive!  There are 5 questions to answer with many expressions to write. 

Solution

A few notes about this problem

1. In this problem, the variable was defined for
you.  Let x represent the number of
children’s tickets sold tells what x stands for in this problem.  If this had not been done for you, you might
have written it like this:

        Let x = the number of children’s tickets sold

2. For the first expression, I knew that 10 more adult
tickets were sold.  Since more means add,
my expression was x +10.  Since the
direction asked for an expression, I don’t need an equal sign.  An equation is written with an equal sign and
an expression is without an equal sign. 
At this point we don’t know the total number of tickets.

3. For the second expression, I knew that my key words,
twice as many meant two times as many. 
So my expression was 2x.

4. 
We know that to find the total
price we have to multiply the price of each ticket by the number of
tickets.  Take note that since x + 10 is the quantity of adult tickets, you must put
it in parentheses!  So, when you multiply
by the price of $5 you have to distribute the 5.

5.  Once I solve
for x, I know the number of children’s tickets and I can take my expressions
that I wrote for #1 and substitute 50 for x to figure out how many adult and
senior tickets were sold.

Example 2 — Identifying a Constant

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

2. How many minutes were charged on this bill?

Solution

A cell phone company charges a monthly rate of $12.95 and $0.25 a minute per call. The bill for m minutes is $21.20.

1. Write an equation that models this situation.

Notes For Example 2

• $12.95 is a monthly rate. Since this is a set fee for each
  month, I know that this is a constant. The rate does not change;
  therefore, it is not associated with a variable.
• $0.25 per
  minute per call requires a variable because the total amount will change
  based on the number of minutes. Therefore, we use the expression 0.25m
• You must solve the equation to determine the value for m, which is the number of minutes charged.

The last example is a word problem that requires an equation with variables on both sides.

Example 3 — Equations with Variables on Both Sides

You have $60 and your sister has $120. You are saving $7 per week and your sister is saving $5 per week. How long will it be before you and your sister have the same amount of money? Write an equation and solve.

Solution

Notes for Example 3

• $60 and $120 are constants because this is the amount of money that they each have to begin with. This amount does not change.
• $7 per week and $5 per week are rates. They key word «per» in this situation means to multiply.
• The key word «same» in this problem means that I am going to set my two expressions equal to each other.
• When we set the two expressions equal, we now have an equation with variables on both sides.
• After solving the equation, you find that x = 30, which means that after 30 weeks, you and your sister will have the same amount of money.

I’m hoping that these three examples will help you as you solve real world problems in Algebra!

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3. Algebra Word Problems

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What are word problems?

A word problem is a math problem written out as a short story or scenario. Basically, it describes a realistic problem and asks you to imagine how you would solve it using math. If you’ve ever taken a math class, you’ve probably solved a word problem. For instance, does this sound familiar?

Johnny has 12 apples. If he gives four to Susie, how many will he have left?

You could solve this problem by looking at the numbers and figuring out what the problem is asking you to do. In this case, you’re supposed to find out how many apples Johnny has left at the end of the problem. By reading the problem, you know Johnny starts out with 12 apples. By the end, he has 4 less because he gave them away. You could write this as:

12 — 4

12 — 4 = 8, so you know Johnny has 8 apples left.

Word problems in algebra

If you were able to solve this problem, you should also be able to solve algebra word problems. Yes, they involve more complicated math, but they use the same basic problem-solving skills as simpler word problems.

You can tackle any word problem by following these five steps:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. Check your work.

We’ll work through an algebra word problem using these steps. Here’s a typical problem:

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took two days, and the van cost $360. How many miles did she drive?

It might seem complicated at first glance, but we already have all of the information we need to solve it. Let’s go through it step by step.

Step 1: Read through the problem carefully.

With any problem, start by reading through the problem. As you’re reading, consider:

• What question is the problem asking?
• What information do you already have?

Let’s take a look at our problem again. What question is the problem asking? In other words, what are you trying to find out?

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

There’s only one question here. We’re trying to find out how many miles Jada drove. Now we need to locate any information that will help us answer this question.

There are a few important things we know that will help us figure out the total mileage Jada drove:

• The van cost $30 per day.
• In addition to paying a daily charge, Jada paid $0.50 per mile.
• Jada had the van for 2 days.
• The total cost was $360.

Step 2: Represent unknown numbers with variables.

In algebra, you represent unknown numbers with letters called variables. (To learn more about variables, see our lesson on reading algebraic expressions.) You can use a variable in the place of any amount you don’t know. Looking at our problem, do you see a quantity we should represent with a variable? It’s often the number we’re trying to find out.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

Since we’re trying to find the total number of miles Jada drove, we’ll represent that amount with a variable—at least until we know it. We’ll use the variable m for miles. Of course, we could use any variable, but m should be easy to remember.

Step 3: Translate the rest of the problem.

Let’s take another look at the problem, with the facts we’ll use to solve it highlighted.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

We know the total cost of the van, and we know that it includes a fee for the number of days, plus another fee for the number of miles. It’s $30 per day, and $0.50 per mile. A simpler way to say this would be:

$30 per day plus $0.50 per mile is $360.

If you look at this sentence and the original problem, you can see that they basically say the same thing: It cost Jada $30 per day and $0.50 per mile, and her total cost was $360. The shorter version will be easier to translate into a mathematical expression.

Let’s start by translating $30 per day. To calculate the cost of something that costs a certain amount per day, you’d multiply the per-day cost by the number of days—in other words, 30 per day could be written as 30 ⋅days, or 30 times the number of days. (Not sure why you’d translate it this way? Check out our lesson on writing algebraic expressions.)

$30 per day and $.50 per mile is $360

$30 ⋅ day + $.50 ⋅ mile = $360

As you can see, there were a few other words we could translate into operators, so and $.50 became + $.50, $.50 per mile became $.50 ⋅ mile, and is became =.

Next, we’ll add in the numbers and variables we already know. We already know the number of days Jada drove, 2, so we can replace that. We’ve also already said we’ll use m to represent the number of miles, so we can replace that too. We should also take the dollar signs off of the money amounts to make them consistent with the other numbers.

$30 ⋅ day + $.50 ⋅ mile = $360

30 ⋅ 2 + .5 ⋅ m = 360

Now we have our expression. All that’s left to do is solve it.

Step 4: Solve the problem.

This problem will take a few steps to solve. (If you’re not sure how to do the math in this section, you might want to review our lesson on simplifying expressions.) First, let’s simplify the expression as much as possible. We can multiply 30 and 2, so let’s go ahead and do that. We can also write .5 ⋅ m as 0.5m.

30 ⋅ 2 + .5 ⋅ m = 360

60 + .5m = 360

Next, we need to do what we can to get the m alone on the left side of the equals sign. Once we do that, we’ll know what m is equal to—in other words, it will let us know the number of miles in our word problem.

We can start by getting rid of the 60 on the left side by subtracting it from both sides.

The only thing left to get rid of is .5. Since it’s being multiplied with m, we’ll do the reverse and divide both sides of the equation with it.

.5m / .5 is m and 300 / 0.50 is 600, so m = 600. In other words, the answer to our problem is 600—we now know Jada drove 600 miles.

Step 5: Check the problem.

To make sure we solved the problem correctly, we should check our work. To do this, we can use the answer we just got—600—and calculate backward to find another of the quantities in our problem. In other words, if our answer for Jada’s distance is correct, we should be able to use it to work backward and find another value, like the total cost. Let’s take another look at the problem.

The rate to rent a small moving van is $30 per day, plus $0.50 per mile. Jada rented a van to drive to her new home. It took 2 days, and the van cost $360. How many miles did she drive?

According to the problem, the van costs $30 per day and $0.50 per mile. If Jada really did drive 600 miles in 2 days, she could calculate the cost like this:

$30 per day and $0.50 per mile

30 ⋅ day + .5 ⋅ mile

30 ⋅ 2 + .5 ⋅ 600

60 + 300

360

According to our math, the van would cost $360, which is exactly what the problem says. This means our solution was correct. We’re done!

While some word problems will be more complicated than others, you can use these basic steps to approach any word problem. On the next page, you can try it for yourself.

Practice!

Let’s practice with a couple more problems. You can solve these problems the same way we solved the first one—just follow the problem-solving steps we covered earlier. For your reference, these steps are:

1. Read through the problem carefully, and figure out what it’s about.
2. Represent unknown numbers with variables.
3. Translate the rest of the problem into a mathematical expression.
4. Solve the problem.
5. Check your work.

If you get stuck, you might want to review the problem on page 1. You can also take a look at our lesson on writing algebraic expressions for some tips on translating written words into math.

Problem 1

Try completing this problem on your own. When you’re done, move on to the next page to check your answer and see an explanation of the steps.

A single ticket to the fair costs $8. A family pass costs $25 more than half of that. How much does a family pass cost?

Problem 2

Here’s another problem to do on your own. As with the last problem, you can find the answer and explanation to this one on the next page.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Problem 1 Answer

Here’s Problem 1:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

Answer: $29

Let’s solve this problem step by step. We’ll solve it the same way we solved the problem on page 1.

Step 1: Read through the problem carefully

The first in solving any word problem is to find out what question the problem is asking you to solve and identify the information that will help you solve it. Let’s look at the problem again. The question is right there in plain sight:

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

So is the information we’ll need to answer the question:

• A single ticket costs $8.
• The family pass costs $25 more than half the price of the single ticket.

Step 2: Represent the unknown numbers with variables

The unknown number in this problem is the cost of the family pass. We’ll represent it with the variable f.

Step 3: Translate the rest of the problem

Let’s look at the problem again. This time, the important facts are highlighted.

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

In other words, we could say that the cost of a family pass equals half of $8, plus $25. To turn this into a problem we can solve, we’ll have to translate it into math. Here’s how:

1. First, replace the cost of a family pass with our variable f.

f equals half of $8 plus $25

2. Next, take out the dollar signs and replace words like plus and equals with operators.

f = half of 8 + 25

3. Finally, translate the rest of the problem. Half of can be written as 1/2 times, or 1/2 ⋅ :

f = 1/2 ⋅ 8 + 25

Step 4: Solve the problem

Now all we have to do is solve our problem. Like with any problem, we can solve this one by following the order of operations.

1. f is already alone on the left side of the equation, so all we have to do is calculate the right side.

f = 1/2 ⋅ 8 + 25

2. First, multiply 1/2 by 8. 1/2 ⋅ 8 is 4.

f = 4 + 25

3. Next, add 4 and 25. 4 + 25 equals 29 .

f = 29

That’s it! f is equal to 29. In other words, the cost of a family pass is $29.

Step 5: Check your work

Finally, let’s check our work by working backward from our answer. In this case, we should be able to correctly calculate the cost of a single ticket by using the cost we calculated for the family pass. Let’s look at the original problem again.

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

We calculated that a family pass costs $29. Our problem says the pass costs $25 more than half the cost of a single ticket. In other words, half the cost of a single ticket will be $25 less than $29.

1. We could translate this into this equation, with s standing for the cost of a single ticket.

1/2s = 29 — 25

2. Let’s work on the right side first. 29 — 25 is 4.

1/2s = 4

3. To find the value of s, we have to get it alone on the left side of the equation. This means getting rid of 1/2. To do this, we’ll multiply each side by the inverse of 1/2: 2.

s = 8

According to our math, s = 8. In other words, if the family pass costs $29, the single ticket will cost $8. Looking at our original problem, that’s correct!

A single ticket to the fair costs $8. A family pass costs $25 more than half that. How much does a family pass cost?

So now we’re sure about the answer to our problem: The cost of a family pass is $29.

Problem 2 Answer

Here’s Problem 2:

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

Answer: $70

Let’s go through this problem one step at a time.

Step 1: Read through the problem carefully

Start by asking what question the problem is asking you to solve and identifying the information that will help you solve it. What’s the question here?

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

To solve the problem, you’ll have to find out how much money Mo gave to charity. All the important information you need is in the problem:

• The amount Flor donated is three times as much the amount Mo donated
• Flor and Mo’s donations add up to $280 total

Step 2: Represent the unknown numbers with variables

The unknown number we’re trying to identify in this problem is Mo’s donation. We’ll represent it with the variable m.

Step 3: Translate the rest of the problem

Here’s the problem again. This time, the important facts are highlighted.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

The important facts of the problem could also be expressed this way:

Mo’s donation plus Flor’s donation equals $280

Because we know that Flor’s donation is three times as much as Mo’s donation, we could go even further and say:

Mo’s donation plus three times Mo’s donation equals $280

We can translate this into a math problem in only a few steps. Here’s how:

1. Because we’ve already said we’ll represent the amount of Mo’s donation with the variable m, let’s start by replacing Mo’s donation with m.

m plus three times m equals $280

2. Next, we can put in mathematical operators in place of certain words. We’ll also take out the dollar sign.

m + three times m = 280

3. Finally, let’s write three times mathematically. Three times m can also be written as 3 ⋅ m, or just 3m.

m + 3m = 280

Step 4: Solve the problem

It will only take a few steps to solve this problem.

1. To get the correct answer, we’ll have to get m alone on one side of the equation.

m + 3m = 280

2. To start, let’s add m and 3m. That’s 4m.

4m = 280

3. We can get rid of the 4 next to the m by dividing both sides by 4. 4m / 4 is m, and 280 / 4 is 70.

m = 70.

We’ve got our answer: m = 70. In other words, Mo donated $70.

Step 5: Check your work

The answer to our problem is $70, but we should check just to be sure. Let’s look at our problem again.

Flor and Mo both donated money to the same charity. Flor gave three times as much as Mo. Between the two of them, they donated $280. How much money did Mo give?

If our answer is correct, $70 and three times $70 should add up to $280.

1. We can write our new equation like this:

70 + 3 ⋅ 70 = 280

2. The order of operations calls for us to multiply first. 3 ⋅ 70 is 210.

70 + 210 = 280

3. The last step is to add 70 and 210. 70 plus 210 equals 280.

280 = 280

280 is the combined cost of the tickets in our original problem. Our answer is correct: Mo gave $70 to charity.

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Related Pages
Basic Algebra
Combining Like Terms
Solving Equations
More Algebra Lessons

Step 1: Translate the problem into equations with variables

First, we need to translate the word problem into equation(s) with variables.
Then, we need to solve the equation(s) to find the solution(s) to the
word problems.

Translating words to equations

How to recognize some common types of algebra word problems and how to solve them step by step.

The following shows how to approach the common types of algebra word problems.

Age Problems usually compare the ages of people.
They may involve a single person, comparing his/her age in the past, present or future.
They may also compare the ages involving more than one person.

Average Problems involve the computations for
arithmetic mean or
weighted average of different quantities.
Another common type of average problems is the average speed computation.

Coin Problems deal with items with denominated values.
Similar word problems are Stamp Problems and
Ticket Problems.

Consecutive Integer Problems deal with consecutive numbers.
The number sequences may be Even or
Odd, or some other simple number sequences.

Digit Problems involve the relationship and manipulation of digits in numbers.

Distance Problems involve the distance an object travels at a rate over a period of time.
We can have objects that Travel at Different Rates, objects that
Travel in Different Directions or we may need to find the distance
Given the Total Time

Fraction Problems involve fractions or parts of a whole.

Geometry Word Problems deal with geometric figures and angles described in words.

This include geometry word problems Involving Perimeters,
Involving Areas and
Involving Angles

Integer Problems involve numerical representations of word problems.
The integer word problems may Involve 2 Unknowns or may
Involve More Than 2 Unknowns

Interest Problems involve calculations of simple interest.
Lever Problems deal with the lever principle described in word problems.
Lever problem may involve 2 Objects or More than 2 Objects

Mixture Problems involve items or quantities of different values that are
mixed together. This involve
Adding to a Solution,
Removing from a Solution,
Replacing a Solution,or Mixing Items of Different Values

Motion Word Problems are word problems that uses the distance, rate and time formula.

Number Sequence Problems use number sequences in the construction of word problems.
You may be asked to find the
Value of a Particular Term or the Pattern of a Sequence

Proportion Problems involve proportional and inversely proportional relationships of various quantities.

Ratio Problems require you to relate quantities of different items in certain
known ratios, or work out the ratios given certain quantities.
This could be Two-Term Ratios or
Three-Term Ratios

Symbol Problems

Variation Word Problems may consist of Direct Variation Problems,
Inverse Variation Problems or Joint Variation Problems

Work Problems involve different people doing work together at different rates.
This may be for Two Persons,
More Than Two Persons or
Pipes Filling up a Tank

For more Algebra Word Problems and Algebra techniques, go to our
Algebra page

Step 2: Solving the equations — finding the values of the variables for the equations

• How to solve equations?

Algebra Word Problems with examples, videos and step-by-step solutions

• Age Word Problems

• Average Word Problems

• Coin Word Problems

• Consecutive Integer Word Problems

• Digit Word Problems

• Distance Word Problems

• Fraction & Percent Word Problems

• Geometry Word Problems

• Integer Word Problems

• Interest Word Problems

• Lever Word Problems

• Mixture Word Problems

• Money & Coin Word Problems

• Motion & Distance Word Problems

• Number Sequence Word Problems

• Proportion Word Problems

• Word Problems using Quadratic Equations

• Ratio Word Problems

• Symbol Word Problems

• Variation Word Problems

• Work Word Problems

• Ways to solve Word Problems

Have a look at the following videos for some introduction of how to solve algebra problems:

Example:
Angela sold eight more new cars this year than Carmen. If together they sold a total of 88 cars, how many cars did each of them sell?

• Show Video Lesson

Example:
One number is 4 times as large as another. Their sum is 45. Find the numbers.

• Show Video Lesson

Example:
Devon is going to make 3 shelves for her father. She has a piece of lumber 12 feet long. She wants
the top shelf to be half a foot shorter than the middle, and the bottom shelf to be half a foot
shorter than twice the length of the top shelf. How long will each shelf be if she uses the entire 12 feet of wood?

• Show Video Lesson

Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own
problem and check your answer with the step-by-step explanations.

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