Learning Objectives
Define a variable with a clear meaning and unit.
Translate a relationship into an expression, equation, or inequality.
Interpret and check the solution against the original situation.
Prerequisites: expressions, one- and two-step equations, and inequalities. A word problem is a description of relationships. Begin by deciding what is unknown and how the given quantities connect.
1. Name the Unknown
Write a sentence such as: let n be the number of notebooks purchased. This is more useful than writing only “let n be the answer.” A clear definition tells you the unit and whether negative or fractional values can make sense.
Description | Translation |
|---|---|
Five more than n | n + 5 |
Three times n | 3n |
Seven less than n | n − 7 |
Seven minus n | 7 − n |
At most 12 | ≤ 12 |
Exactly 12 | = 12 |
Read the whole relationship rather than searching for one keyword. “Less than” can appear in a subtraction phrase or in an inequality statement. For example, “seven less than n” is n − 7, while “n is less than seven” is n < 7.
2. Separate Fixed and Repeated Amounts
Worked Example: A Total Cost
A craft activity costs a fixed $5 entry fee plus $3 per item made. The total bill is $20. Let n be the number of items.
The repeated cost is 3n, so the equation is 5 + 3n = 20.
Subtract 5 and divide by 3: n = 5. Check: the entry fee plus five item charges is 5 + 15 = $20. Five is a valid nonnegative whole-number count.
3. Use a Formula for a Geometric Relationship
Worked Example: Perimeter
A rectangle has width w centimetres and length w + 3 centimetres. Its perimeter is 30 cm. A perimeter adds all four side lengths.
Write 2w + 2(w + 3) = 30. Distribute and combine to get 4w + 6 = 30.
Then 4w = 24 and w = 6. The length is 9 cm. Check: 2(6) + 2(9) = 30 cm.
The width variable does not automatically answer every part of the question. If the problem also asks for area, use both dimensions: 6 × 9 = 54 cm². Perimeter and area have different units.
4. Limits Become Inequalities
Worked Example: Staying within a Budget
You have $29. Delivery costs $5, and each notebook costs $4. Let n count notebooks. Spending at most $29 gives 5 + 4n ≤ 29.
Subtract 5 and divide by 4 to obtain n ≤ 6.
The greatest possible count is 6. This uses all $29, which is permitted by “at most.” If the problem required spending less than $29, six would be excluded.
5. Check the Situation, Not Just the Algebra
After solving, ask whether the units agree, whether a count is a whole number, and whether the result satisfies the stated relationship. A negative length or a fractional number of indivisible objects can reveal that an equation was set up incorrectly or that the given conditions cannot all hold.
Common Mistakes
Preserve the order of subtraction phrases.
Do not multiply a one-time fee by the number of items.
Write a final sentence naming the quantity found and its unit.
6. Practice
1. A number increased by 8 equals 21. Find the number.
2. A rental costs $7 plus $4 per hour. The bill is $31. How many hours were rented?
3. A rectangle has length 10 cm and perimeter 34 cm. Find its width.
4. A $2 entrance charge and $5 per ride must total less than $23. What is the maximum whole number of rides?
Worked Solutions
1. n + 8 = 21 gives n = 13.
2. 7 + 4h = 31 gives 4h = 24 and h = 6 hours.
3. 2(10) + 2w = 34 gives 2w = 14 and w = 7 cm.
4. 2 + 5n < 23 gives n < 21/5 = 4.2. The largest whole number is 4. Its cost is $22, which is less than $23.
7. Summary
- Define the variable before translating the relationship.
- Separate one-time quantities from repeated quantities.
- Choose equality or an inequality according to the wording.
- Interpret the solution with units and contextual restrictions.