Use a Problem Solving Strategy

Approach word problems with a positive attitude

The world is full of word problems: How much money do I need to fill the car with gas? How much should I tip the server at a restaurant? How many socks should I pack for vacation? Have you ever had negative thoughts about word problems, like “I don’t understand word problems” or “I don’t know whether to add, subtract, multiply, or divide”?

Now that we can solve equations, we are ready to apply our new skills to word problems. Think of something that you can do now but couldn’t do three years ago. Whether it’s driving a car, cooking a meal, or speaking a new language, you have been able to learn and master a new skill. Word problems are no different — even if you have struggled with them in the past, you have acquired many new math skills that will help you succeed now. Start with a fresh slate: “I am better prepared now. I think I can.”

Use a problem-solving strategy for word problems

In earlier chapters, you translated word phrases into algebraic expressions and word sentences into algebraic equations, and solved word problems as long as the situation was familiar and the math was not too complicated. Now we’ll develop a strategy you can use to solve any word problem.

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

Step 1. Read the problem. Make sure you understand all the words and ideas — you may need to read it two or more times.

Step 2. Identify what you are looking for: the words “what was the original price of the shirt” tell us we’re looking for the original price of the shirt.

Step 3. Name what you are looking for. Choose a variable to represent that quantity. Let $p =$ the original price of the shirt.

Step 4. Translate. Restate the problem in one sentence with all the important information, then translate the sentence into an equation: “$18$ is one-half of the original price” becomes

Step 5. Solve the equation using good algebra techniques:

Step 6. Check the answer in the problem and make sure it makes sense: we found $p = 36$, so the original price was $\text{\textdollar}36$. Does that make sense? Yes — $18$ is one-half of $36$, and the shirt was on sale at half the original price.

Step 7. Answer the question with a complete sentence: the original price of the shirt was $\text{\textdollar}36$.

Let’s use the strategy on a problem with more than one unknown quantity.

Example. Yash brought apples and bananas to a picnic. The number of apples was three more than twice the number of bananas. Yash brought $11$ apples to the picnic. How many bananas did he bring?

We identify what we’re looking for: the number of bananas. Let $b =$ the number of bananas. Translating “the number of apples was three more than twice the number of bananas” with $11$ apples gives

Solving: subtract $3$ from each side to get $8 = 2b$, then divide each side by $2$ to get $4 = b$. Checking: twice $4$ bananas is $8$, and three more than $8$ is $11$ — that matches the $11$ apples. Yash brought $4$ bananas to the picnic.

In Solve Sales Tax, Commission, and Discount Applications, we learned how to translate and solve basic percent equations. Here’s another application of that skill within the same seven-step strategy.

Example. Nga’s car insurance premium increased by $\text{\textdollar}60$, which was $8\%$ of the original cost. What was the original cost of the premium?

Let $c =$ the original cost. Translating “$\text{\textdollar}60$ was $8\%$ of the original cost” gives $60 = 0.08c$. Dividing both sides by $0.08$ gives $c = 750$. Checking: is $8\%$ of $750$ equal to $60$? Yes, $0.08(750) = 60$. The original cost of Nga’s premium was $\text{\textdollar}750$.

Solve number problems

Now we will translate and solve number problems. In number problems you are given clues about one or more numbers, and you use these clues to build an equation. Number problems don’t usually arise on an everyday basis, but they provide good practice with the problem-solving strategy. Watch for clue words such as difference, of, and and.

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

Let $n =$ the number. “The difference of a number and $6$ is $13$” translates to $n - 6 = 13$. Adding $6$ to both sides gives $n = 19$. Checking: the difference of $19$ and $6$ is $13$. It checks. The number is $19$.

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

Let $n =$ the number. “The sum of twice a number and $7$ is $15$” translates to $2n + 7 = 15$. Subtracting $7$ from each side gives $2n = 8$, and dividing each side by $2$ gives $n = 4$. Checking: twice $4$ is $8$, and $8 + 7 = 15$. The number is $4$.

Some number problems ask you to find two or more numbers. It’s tempting to name them all with different variables, but we only know how to solve equations with one variable — so we define the numbers in terms of the same variable.

Example. One number is five more than another. The sum of the numbers is twenty-one. Find the numbers.

Let $n =$ the first number. “One number is five more than another” tells us the second number is $n + 5$. “The sum of the numbers is $21$” translates to

Combining like terms: $2n + 5 = 21$. Subtracting $5$ from both sides and simplifying: $2n = 16$, so $n = 8$ is the first number, and $n + 5 = 13$ is the second. Checking: is $13$ five more than $8$? Yes. Is their sum $21$? Yes, $8 + 13 = 21$. The numbers are $8$ and $13$.

Some pairs of numbers are related by subtraction instead of addition, and some involve negative sums — the same strategy still applies.

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 $n - 4$ is the second number (since it’s four less). “The sum of two numbers is negative fourteen” translates to

Combining like terms: $2n - 4 = -14$. Adding $4$ to each side: $2n = -10$, so $n = -5$ is the first number, and $n - 4 = -9$ is the second. Checking: is $-9$ four less than $-5$? Yes. Is their sum $-14$? Yes, $-5 + (-9) = -14$. The numbers are $-5$ and $-9$.

Number problems can also relate one number to twice another.

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

Let $x =$ the first number, so $2x + 10$ is the second number. “Their sum is one” translates to

Combining like terms: $3x + 10 = 1$. Subtracting $10$ from each side gives $3x = -9$, so $x = -3$ is the first number. Substituting to get the second: $2(-3) + 10 = 4$. Checking: is $4$ ten more than twice $-3$? Yes, $2(-3) + 10 = 4$. Is their sum $1$? Yes, $-3 + 4 = 1$. The numbers are $-3$ and $4$.

Consecutive integers are integers that immediately follow each other, like $1, 2, 3, 4, \dots$ or $-10, -9, -8, -7, \dots$. Each number is one more than the one before it, so if we call the first integer $n$, the next consecutive integer is $n + 1$, and the one after that is $n + 2$.

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

Let $n =$ the first integer, so $n + 1$ is the next consecutive integer. “The sum of the integers is $47$” translates to $n + (n + 1) = 47$. Combining like terms: $2n + 1 = 47$. Subtracting $1$ from each side gives $2n = 46$, so $n = 23$ is the first integer, and $n + 1 = 24$ is the second. Checking: $23 + 24 = 47$. The two consecutive integers are $23$ and $24$.

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

Let $n =$ the first integer, so $n + 1$ is the second consecutive integer and $n + 2$ is the third. “The sum of the three integers is $42$” translates to

Combining like terms: $3n + 3 = 42$. Subtracting $3$ from each side gives $3n = 39$, so $n = 13$ is the first integer. Then $n + 1 = 14$ is the second, and $n + 2 = 15$ is the third. Checking: $13 + 14 + 15 = 42$. The three consecutive integers are $13$, $14$, and $15$.

Key terms

problem-solving strategy — the seven-step method (read, identify, name, translate, solve, check, answer) used to set up and solve any word problem. number problem — a word problem giving clues about one or more numbers, solved by naming the numbers in terms of a single variable and translating the clues into an equation. consecutive integers — integers that immediately follow each other, each one more than the number before it.

This section is adapted from Prealgebra 2e, Section 9.1: 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: omitted the Be Prepared quiz, the illustrative figures of a student’s negative and positive thoughts, the Links to Literacy callout, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.