Table of Contents
Turning everyday situations into equations
In this chapter, we focus on word problems that lead to one-variable linear equations. You should already know how to solve such equations; here we practice how to set them up from a description in words.
The key skill is translation: turning a situation described in everyday language into a mathematical equation.
General strategy for word problems
When you read a word problem, do not jump straight to numbers. Follow a clear process:
- Read the problem carefully
- Read the whole problem once without doing any calculations.
- Identify what is being asked: what do you need to find?
- Choose a variable
- Pick a letter (often $x$) to represent the unknown quantity.
- Clearly state what your variable means in words.
- Translate relationships into an equation
- Look for phrases that describe operations: addition, subtraction, multiplication, division.
- Use your variable and the given numbers to write an equation that matches the relationships in the problem.
- Solve the equation
- Use methods from one-variable equations (handled in the previous chapter).
- Answer the question in words
- Translate your solution back into the context of the problem.
- Include units (dollars, meters, years, etc.).
- Check your answer
- Substitute your answer back into the original situation and see if it makes sense.
- Check that it answers exactly what was asked.
Common language-to-equation translations
Many word problems use similar phrases. Learn how they usually translate:
- Addition phrases (often become “$+$”):
- “more than”
- “increased by”
- “added to”
- “the sum of”
Example: “5 more than a number” $\rightarrow$ if the number is $x$, then it is $x + 5$.
- Subtraction phrases (often become “$-$”):
- “less than”
- “decreased by”
- “minus”
- “the difference between”
Be careful with order:
- “5 less than a number” $\rightarrow$ $x - 5$
- “a number less 5” $\rightarrow$ $x - 5$
- “the difference between 5 and a number” $\rightarrow$ $5 - x$
- Multiplication phrases (often become “$\times$”):
- “times”
- “twice” (2 times)
- “triple” (3 times)
- “of” (in some contexts, like “half of,” “3 times as many as”)
Examples:
- “twice a number” $\rightarrow$ $2x$
- “3 times as many as a number” $\rightarrow$ $3x$
- Division phrases (often become “$\div$” or fractions):
- “per”
- “out of”
- “divided by”
- “split equally”
Examples:
- “a number divided by 4” $\rightarrow$ $\dfrac{x}{4}$
- “half of a number” $\rightarrow$ $\dfrac{1}{2}x$
- Equality phrases (often become “$=$”):
- “is”
- “are”
- “gives”
- “results in”
- “equals”
Example: “Twice a number is 10” $\rightarrow$ $2x = 10$
One-step word problems
These problems need only one operation to form the equation.
One-step addition and subtraction
Example 1 (addition):
A child has some stickers. After receiving 7 more stickers, she has 15 stickers. How many stickers did she have at first?
- Let $x$ = number of stickers she had at first.
- “After receiving 7 more” $\rightarrow$ $x + 7$.
- “She has 15 stickers” $\rightarrow$ $x + 7 = 15$.
- Solve (using earlier knowledge): $x = 8$.
- Answer: She had 8 stickers at first.
Example 2 (subtraction):
A box originally contained some apples. After 9 apples were taken out, there are 5 apples left. How many apples were originally in the box?
- Let $x$ = number of apples originally in the box.
- “After 9 apples were taken out” $\rightarrow$ $x - 9$.
- “There are 5 apples left” $\rightarrow$ $x - 9 = 5$.
- Solving gives $x = 14$.
- Answer: There were 14 apples originally.
One-step multiplication and division
Example 3 (multiplication):
A number is tripled to give 27. What is the number?
- Let $x$ = the number.
- “Tripled” $\rightarrow$ $3x$.
- “Gives 27” $\rightarrow$ $3x = 27$.
- Solving gives $x = 9$.
- Answer: The number is 9.
Example 4 (division):
When a number is divided by 4, the result is 7. What is the number?
- Let $x$ = the number.
- “Divided by 4” $\rightarrow$ $\dfrac{x}{4}$.
- “The result is 7” $\rightarrow$ $\dfrac{x}{4} = 7$.
- Solving gives $x = 28$.
- Answer: The number is 28.
Two-step word problems
These problems usually involve two operations. You will need to form and solve simple linear equations with more than one step.
Typical pattern: “Multiply, then add” or “Multiply, then subtract”
Example 5:
A number is doubled and then 5 is added. The result is 19. What is the number?
- Let $x$ = the number.
- “Doubled” $\rightarrow$ $2x$.
- “Then 5 is added” $\rightarrow$ $2x + 5$.
- “The result is 19” $\rightarrow$ $2x + 5 = 19$.
- Solve: $2x + 5 = 19$ leads to $x = 7$.
- Answer: The number is 7.
Example 6:
Three times a number minus 4 is 17. Find the number.
- Let $x$ = the number.
- “Three times a number” $\rightarrow$ $3x$.
- “Minus 4” $\rightarrow$ $3x - 4$.
- “Is 17” $\rightarrow$ $3x - 4 = 17$.
- Solve: $3x - 4 = 17$ leads to $x = 7$.
- Answer: The number is 7.
Using parentheses when needed (though solving is not the focus)
Sometimes, the wording means “do something to the whole expression.” Parentheses are used here.
Example 7:
If 4 is added to a number and the result is then doubled, the final answer is 20. What is the number?
- Let $x$ = the number.
- “4 is added to a number” $\rightarrow$ $x + 4$.
- “The result is then doubled” $\rightarrow$ $2(x + 4)$.
- “The final answer is 20” $\rightarrow$ $2(x + 4) = 20$.
- Solve using your equation skills to find $x = 6$.
- Answer: The number is 6.
Word problems about money and cost
Many practical problems involve buying items, total costs, and fixed fees.
Typical features:
- A fixed cost (does not depend on quantity).
- A variable cost (depends on how many items or hours).
These usually become linear equations.
Cost problems with a fixed fee
Example 8:
A taxi company charges a fixed fee of \$3 plus \$2 per kilometer. If a trip costs \$15, how many kilometers was the trip?
- Let $x$ = number of kilometers.
- “Fixed fee of \$3 plus \$2 per kilometer” $\rightarrow$ $3 + 2x$.
- “Trip costs \$15” $\rightarrow$ $3 + 2x = 15$.
- Solve: $3 + 2x = 15$ gives $x = 6$.
- Answer: The trip was 6 kilometers.
Buying multiple items
Example 9:
Notebooks cost \$4 each. A student buys some notebooks and spends \$28 in total. How many notebooks did the student buy?
- Let $x$ = number of notebooks.
- “\$4 each” $\rightarrow$ cost is $4x$.
- “Spends \$28 in total” $\rightarrow$ $4x = 28$.
- Solve for $x$ to get $x = 7$.
- Answer: The student bought 7 notebooks.
Example 10 (fixed cost plus items):
A phone plan costs a fixed \$10 per month plus \$0.50 for each text message. One month, the bill is \$25. How many text messages were sent?
- Let $x$ = number of text messages.
- “Fixed \$10 plus \$0.50 for each text” $\rightarrow$ $10 + 0.5x$.
- “Bill is \$25” $\rightarrow$ $10 + 0.5x = 25$.
- Solve to find $x = 30$.
- Answer: 30 text messages were sent.
Word problems about age
Age problems often compare ages now or at different times (in the past or future).
Ages now
Example 11:
Maria is 4 years older than her brother Alex. If Alex is $x$ years old, how old is Maria?
- Let $x$ = Alex’s age (this is given as an expression).
- “Maria is 4 years older” $\rightarrow$ Maria’s age is $x + 4$.
If instead the problem stated:
Maria is 4 years older than her brother Alex. Together, their ages add up to 28. How old is Alex?
- Let $x$ = Alex’s age now.
- Maria’s age: $x + 4$.
- “Together, their ages add up to 28”:
$$x + (x + 4) = 28.$$ - Solve for $x$ to find Alex’s age, then add 4 for Maria’s.
Ages in the past or future
Example 12:
In 5 years, John will be 3 times as old as he is now. How old is he now?
- Let $x$ = John’s current age.
- “In 5 years” $\rightarrow$ $x + 5$.
- “Will be 3 times as old as he is now” $\rightarrow$ $3x$.
- Equation: $$x + 5 = 3x.$$
- Solve for $x$.
- Answer in years.
Example 13:
Five years ago, Emma was 7 years younger than she is now. If she is $x$ years old now, express her age 5 years ago.
- Let $x$ = Emma’s age now.
- “Five years ago” $\rightarrow$ $x - 5$.
- This already represents her age 5 years ago, no equation needed unless more information is given.
Word problems involving distance, speed, and time
For these problems, the basic relationship is:
$$\text{distance} = \text{speed} \times \text{time}.$$
Depending on what is unknown, you may rewrite this as needed.
Basic travel problem
Example 14:
A car travels at 60 km per hour for $x$ hours and covers 180 km. How many hours did the car travel?
- Let $x$ = time in hours.
- Use the relationship: distance $=$ speed $\times$ time.
- Distance: 180 km.
- Speed: 60 km per hour.
- Equation: $$60x = 180.$$
- Solve for $x$ to find the travel time.
Going and returning
Example 15:
You walk from home to school at 4 km/h and it takes you 30 minutes. How far is the school from home?
- Let $x$ = distance from home to school in km.
- Time: 30 minutes $= 0.5$ hours.
- Use: distance $=$ speed $\times$ time.
- Equation: $$x = 4 \times 0.5.$$
- Calculate $x$ to find the distance.
Word problems about totals and comparisons
These problems use phrases like “together,” “in total,” “difference,” or “more than/less than.”
Totals (sums)
Example 16:
Two numbers add up to 50. One number is 12 more than the other. Find the numbers.
- Let $x$ = the smaller number.
- The larger number: “12 more than the other” $\rightarrow$ $x + 12$.
- “Add up to 50”:
$$x + (x + 12) = 50.$$ - Solve for $x$, then find $x + 12$.
Differences (comparisons)
Example 17:
The difference between a number and 9 is 20. Find the number.
- Let $x$ = the number.
- “Difference between a number and 9” (number minus 9) $\rightarrow$ $x - 9$.
- “Is 20” $\rightarrow$ $x - 9 = 20$.
- Solve for $x$.
Example 18 (be careful with order):
The difference between 40 and a number is 12. Find the number.
- Let $x$ = the number.
- “Difference between 40 and a number” (40 minus the number) $\rightarrow$ $40 - x$.
- “Is 12” $\rightarrow$ $40 - x = 12$.
- Solve for $x$.
Word problems with equal expressions
Sometimes two different descriptions of the same quantity are set equal.
Example 19:
A movie theater earns the same amount from adult and child tickets for a show. Adult tickets cost \$10 each, and child tickets cost \$6 each. If 50 child tickets are sold, how many adult tickets must be sold to earn the same amount?
- Let $x$ = number of adult tickets sold.
- Money from adult tickets: $10x$.
- Money from child tickets: $6 \times 50 = 300$.
- “Earns the same amount”:
$x = 300.$$ - Solve for $x$.
Example 20:
The perimeter of a rectangle is equal to the perimeter of a square. The rectangle has length 8 cm and width $x$ cm. The square has side length 5 cm. Find $x$.
- Rectangle perimeter: $2(8 + x)$.
- Square perimeter: $4 \times 5 = 20$.
- “Is equal to”:
$(8 + x) = 20.$$ - Solve for $x$ to find the width.
Checking reasonableness of answers
After solving any word problem:
- Substitute the value of your variable back into the original situation.
- See if all conditions are satisfied.
- Think about whether the answer makes sense:
- Are you getting a negative number where only positive makes sense (like age or number of objects)?
- Does the size of the answer seem reasonable compared to the numbers in the problem?
Example 21:
If you get that someone’s age is $-3$ years, you know something is wrong. Re-check:
- Did you choose the variable correctly?
- Did you translate the language into an equation correctly?
- Did you solve the equation correctly?
Practice structures to focus on
When creating or practicing word problems, aim for these structures:
- One-step:
- $x + a = b$
- $x - a = b$
- $ax = b$
- $\dfrac{x}{a} = b$
- Two-step:
- $ax + b = c$
- $ax - b = c$
- $a(x + b) = c$
In each case, practice:
- Defining the variable in words.
- Writing a sentence that matches the equation.
- Writing an equation that matches a sentence.
By concentrating on the translation between words and equations, you develop the central skill for handling linear equation word problems in pre-algebra.