Use a Problem-Solving Strategy

Approach Word Problems with a Positive Attitude

“If you think you can… or think you can’t… you’re right.” —Henry Ford

The world is full of word problems: Will my income qualify me to rent that apartment? How much punch do I need to make for the party? What size diamond can I afford to buy my girlfriend? Should I fly or drive to my family reunion? How much money do I need to fill the car with gas? How much tip should I leave at a restaurant?

Now that we can solve equations, we are ready to apply our new skills to word problems. Many students have had negative experiences with word problems in the past: “I don’t understand word problems,” “My teachers never explained this,” “I don’t know whether to add, subtract, multiply, or divide,” “If I just skip all the word problems, I can probably still pass the class,” “I just can’t do this.”

When we feel we have no control, and continue repeating negative thoughts, we set up barriers to success. We need to calm our fears and change our negative feelings.

Start with a fresh slate and begin to think positive thoughts. If we take control and believe we can be successful, we will be able to master word problems: “I am better prepared now. I think I will begin to understand word problems.” “While word problems were hard in the past, I think I can try them now.” “I think I can! I think I can!” “It may take time, but I can begin to solve word problems.”

Think of something, outside of school, that you can do now but couldn’t do three years ago. Is it driving a car? Snowboarding? Cooking a gourmet meal? Speaking a new language? Your past experiences with word problems happened when you were younger — now you’re older and ready to succeed!

Use a Problem-Solving Strategy for Word Problems

We have reviewed translating English phrases into algebraic expressions, using some basic mathematical vocabulary and symbols. We have also translated English sentences into algebraic equations and solved some word problems. The word problems applied math to everyday situations. We restated the situation in one sentence, assigned a variable, and then wrote an equation to solve the problem. This method works as long as the situation is familiar and the math is not too complicated.

Now we’ll expand our strategy so we can use it to successfully solve any word problem.

Example. Pilar bought a purse on sale for $18, which is one-half of the original price. What was the original price of the purse?

Step 1. Read the problem. Read it two or more times if necessary. Look up any unfamiliar words in a dictionary or on the internet. Is every word familiar?

Step 2. Identify what we are looking for. Read the problem again and look for words that tell us what we are looking for. Here, the words “what was the original price of the purse” tell us what we need to find.

Step 3. Name what we are looking for. Choose a variable to represent the quantity — one that makes it easy to remember what it represents. Let $p =$ the original price of the purse.

Step 4. Translate into an equation. Reread the problem carefully to see how the given information is related, and restate it in one sentence with all the important information:

Translate the sentence into an equation:

Step 5. Solve the equation. Even when the solution seems obvious, using good algebraic techniques prepares you for problems that don’t have such obvious answers.

Step 6. Check the answer in the problem, making sure it makes sense. We found $p = 36$, so the original price was $36. Does that make sense? Yes — 18 is one-half of 36, and the purse was on sale at half the original price.

Step 7. Answer the question with a complete sentence: the original price of the purse was $36.

Let’s try this approach with another example, where the given quantities relate to each other rather than to a single unknown fraction.

Example. Ginny and her classmates formed a study group. The number of girls in the study group was three more than twice the number of boys. There were 11 girls in the study group. How many boys were in the study group?

Step 1. Read the problem.

Step 2. Identify what we are looking for: how many boys were in the study group?

Step 3. Name what we are looking for. Let $b =$ the number of boys.

Step 4. Translate. Restate the problem in one sentence with all the important information: the number of girls (11) was three more than twice the number of boys. Translate into an equation:

Step 5. Solve the equation.

Step 6. Check. First, is our answer reasonable? Yes — having 4 boys in a study group seems fine. The problem says the number of girls was 3 more than twice the number of boys. If there are four boys, does that make eleven girls? Twice 4 boys is 8; three more than 8 is 11. It checks!

Step 7. Answer the question: there were 4 boys in the study group.

Solve Number Problems

Now that we have a problem-solving strategy, we will use it on several different types of word problems. The first type we will work on is “number problems.” Number problems give some clues about one or more numbers. We use these clues to write an equation. Number problems don’t usually arise on an everyday basis, but they provide a good introduction to practicing the problem-solving strategy outlined above.

Example. The difference of a number and six is 13. Find the number.

Step 1. Read the problem. Step 2. Identify what we are looking for: the number. Step 3. Name what we are looking for. Let $n =$ the number.

Step 4. Translate. Look for clue words like “difference… of… and…”. Restate the problem as one sentence: the difference of the number and 6 is 13. Translate into an equation:

Step 5. Solve the equation: $n = 19$.

Step 6. Check. The difference of 19 and 6 is 13. It checks!

Step 7. Answer the question: the number is 19.

Example. The sum of twice a number and seven is 15. Find the number.

Step 1–3. We are looking for the number. Let $n =$ the number.

Step 4. Translate. The sum of twice a number and 7 is 15:

Step 5. Solve.

Step 6. Check. Is the sum of twice 4 and 7 equal to 15? $2 \cdot 4 + 7 = 15$, and $15 = 15$. It checks!

Step 7. Answer. The number is 4.

Did you notice that some of the steps were left out while solving this equation? If you’re not yet ready to leave out these steps, write down as many as you need.

Some number word problems ask us to find two or more numbers. It may be tempting to name them all with different variables, but so far we have only solved equations with one variable. In order to avoid using more than one variable, we will define the numbers in terms of the same variable. Be sure to read the problem carefully to discover how all the numbers relate to each other.

Example. One number is five more than another. The sum of the numbers is 21. Find the numbers.

Step 1. Read the problem. Step 2. Identify what we are looking for: we are looking for two numbers. Step 3. Name. We have two numbers to name, and need a name for each. Let $n =$ the first number. Since the second number is five more than the first, $n + 5 =$ the second number.

Step 4. Translate. Restate the problem as one sentence with all the important information: the sum of the first number and the second number is 21. Translate into an equation, then substitute the variable expressions:

Step 5. Solve the equation.

Find the second number too: $n + 5 = 8 + 5 = 13$.

Step 6. Check. Is one number 5 more than the other? Is thirteen 5 more than 8? Yes. Is the sum of the two numbers 21? $8 + 13 = 21$. Yes, it checks!

Step 7. Answer. The numbers are 8 and 13.

Example. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Let $n =$ the first number, so $n - 4 =$ the second number (one number is four less than the other).

Translate. The sum of the two numbers is negative fourteen:

Solve.

The second number is $n - 4 = -5 - 4 = -9$.

Check. Is $-9$ four less than $-5$? $-5 - 4 = -9$. Yes. Is their sum $-14$? $-5 + (-9) = -14$. Yes, it checks!

Answer. The numbers are $-5$ and $-9$.

Example. One number is ten more than twice another. Their sum is one. Find the numbers.

Let $x =$ the first number, so $2x + 10 =$ the second number (ten more than twice another).

Translate. Their sum is one:

Solve.

The second number is $2x + 10 = 2(-3) + 10 = 4$.

Check. Is ten more than twice $-3$ equal to 4? $2(-3) + 10 = 4$. Yes. Is their sum 1? $-3 + 4 = 1$. Yes, it checks!

Answer. The numbers are $-3$ and $4$.

Consecutive Integers

Some number problems involve consecutive integers. Consecutive integers are integers that immediately follow each other. Examples of consecutive integers are:

Notice that each number is one more than the number preceding it. So if we define the first integer as $n$, the next consecutive integer is $n + 1$. The one after that is one more than $n + 1$, so it is $n + 1 + 1$, which is $n + 2$:

Example. The sum of two consecutive integers is 47. Find the numbers.

Step 1. Read the problem. Step 2. Identify what we are looking for: two consecutive integers. Step 3. Name each number. Let $n =$ the first integer, so $n + 1 =$ the next consecutive integer.

Step 4. Translate. Restate as one sentence: the sum of the integers is 47.

Step 5. Solve the equation.

The next consecutive integer is $n + 1 = 23 + 1 = 24$.

Step 6. Check. $23 + 24 = 47$. It checks!

Step 7. Answer. The two consecutive integers are 23 and 24.

Example. Find three consecutive integers whose sum is $-42$.

Let $n =$ the first integer. Then $n + 1 =$ the second consecutive integer, and $n + 2 =$ the third consecutive integer.

Translate. The sum of the three integers is $-42$:

Solve.

The second integer is $n + 1 = -15 + 1 = -14$. The third integer is $n + 2 = -15 + 2 = -13$.

Check. $-13 + (-14) + (-15) = -42$. It checks!

Answer. The three consecutive integers are $-13$, $-14$, and $-15$.

Consecutive Even and Odd Integers

Now that we have worked with consecutive integers, we will expand our work to include consecutive even integers and consecutive odd integers. Consecutive even integers are even integers that immediately follow one another. Examples of consecutive even integers are:

Notice each integer is 2 more than the number preceding it. If we call the first one $n$, then the next one is $n + 2$. The next one would be $n + 2 + 2$, or $n + 4$:

Consecutive odd integers are odd integers that immediately follow one another. Consider the consecutive odd integers 77, 79, and 81:

Does it seem strange to add 2 (an even number) to get from one odd integer to the next? Whether the problem asks for consecutive even numbers or odd numbers, you don’t have to do anything different — the pattern is still the same: to get from one odd or one even integer to the next, add 2.

Example. Find three consecutive even integers whose sum is 84.

Let $n =$ the first even integer, so $n + 2 =$ the second consecutive even integer, and $n + 4 =$ the third consecutive even integer.

Translate. The sum of the three even integers is 84:

Solve.

The second integer is $n + 2 = 26 + 2 = 28$. The third integer is $n + 4 = 26 + 4 = 30$.

Check. $26 + 28 + 30 = 84$. It checks!

Answer. The three consecutive integers are 26, 28, and 30.

Example. A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

Step 1. Read the problem. Step 2. Identify what we are looking for: how much does the husband earn? Step 3. Name. Let $h =$ the amount the husband earns. Since the wife earns $16,000 less than twice that, $2h - 16{,}000 =$ the amount the wife earns.

Step 4. Translate. Restate the problem in one sentence with all the important information: the amount the husband earns plus the amount the wife earns is $110,000. Translate into an equation:

Step 5. Solve the equation.

The amount the wife earns is $2h - 16{,}000 = 2(42{,}000) - 16{,}000 = 68{,}000$.

Step 6. Check. If the wife earns $68,000 and the husband earns $42,000, is the total $110,000? Yes!

Step 7. Answer. The husband earns $42,000 a year.

Key terms

solution of an equation — a value of a variable that makes a true statement when substituted into the equation. consecutive integers — integers that immediately follow each other, each one more than the previous ($n$, $n+1$, $n+2$, …). consecutive even integers — even integers that immediately follow each other, each two more than the previous ($n$, $n+2$, $n+4$, …). consecutive odd integers — odd integers that immediately follow each other, following the same “add two” pattern as consecutive even integers.

This section is adapted from Elementary Algebra 2e, Section 3.1: Use a Problem-Solving Strategy by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the negative/positive self-talk thought-bubble figures as prose, condensed the seven-step worked examples’ two-column tables into labeled steps, and omitted the Be Prepared quiz, Self Check checklist, media links, and end-of-section exercises; converted the practice problems (“Try Its”) into interactive exercises with instant feedback.