2.5 Word problems into equations

Many algebra questions do not hand you an equation. They describe a situation, and you have to build the equation yourself. The arithmetic is often short; the difficult part is deciding what a variable means and which quantities belong together. A clear translation also makes it easier to notice when a numerical answer is impossible in the original situation.

This is foundational to intermediate algebra. Learn to simplify expressions in Module 2.1 and solve linear equations in Module 2.2 first. OpenStax's models and applications offers more word problems.

A basic idea: name the missing quantity

Suppose a service charges $12$ units to start and $4$ units for each item. If $n$ is the number of items, the total cost is $12+4n$. The $12$ occurs once; the $4$ occurs for every item. If the total is $56$, then $12+4n=56$. This equation has a meaning before we solve it.

Write the variable and its units first: “$n=$ number of items, a nonnegative whole number.” A letter without a definition is easy to confuse with a price, a total, or a count.

A quick tour of translation

1. Identify the question. What quantity must the answer give? Name it and state its units.
2. List the known relationships. Write each in a short sentence or table before doing algebra.
3. Build an equation. Each side should describe the same quantity in the same units.
4. Solve, then check in words. Substitute into the original relationships and apply restrictions such as a positive length or a whole-number count.

For “$x$ is five less than $y$,” write $x=y-5$; word order alone is unreliable. A small test can help: if $y=10$, then $x$ should be $5$.

When two quantities are related, try defining only one variable. If a rectangle's length is $3$ more than its width $w$, write the length as $w+3$. If two independent unknowns remain, use a system as in Module 2.2. When a question says “at least” or “at most,” it creates an inequality rather than an equation; Module 2.4 shows how to solve it.

Application 1: a fixed fee and a per-item fee

A service charges a $12$-unit setup fee and $4$ units per item. A bill is $56$ units. How many items were processed?

1. Name the count. Let $n$ be the number of items, with $n$ a nonnegative whole number.
2. Describe the bill. Setup contributes $12$ units; items contribute $4n$ units. Thus $12+4n=56$.
3. Solve. $4n=44$, so $n=11$.
4. Check the story. Eleven items cost $4(11)=44$ units; with setup, the bill is $56$.

Dividing the entire $56$ by $4$ would incorrectly charge the setup fee for items.

Application 2: a rectangle from its perimeter

A rectangle's length is $3$ units greater than its width. Its perimeter is $34$ units. Find the side lengths and area.

1. Choose one variable. Let width be $w>0$. Length is $w+3$.
2. Use the perimeter relationship. Two widths and two lengths make the perimeter: $2w+2(w+3)=34$.
3. Solve and check. $4w+6=34$, so $w=7$ and length is $10$. The perimeter is $2(7)+2(10)=34$.
4. Answer the extra question. Area is $7\cdot10=70$ square units.

The words “greater than” made the length $w+3$, while “perimeter” required two copies of each side. Checking both parts prevents an equation that fits the numbers but misreads the shape.