Word Problems with Linear Equations

Learning goals

Name one unknown with its units, and build everything from it
Turn the key relationship into a single equation
Write consecutive integers as $n$ , $n + 1$ , $n + 2$
Apply $d = rt$ , total value, the perimeter formula, and percent relationships
Check against the words, and report what was actually asked

Reading a problem as an equation

You already know how to solve a linear equation once you have one. What a word problem asks of you first is to build that equation out of a paragraph, and that is a skill of its own. From the very first lesson of this subject you have been translating phrases into symbols. Here “more than” adds, “less than” subtracts, “times” multiplies, and the verb “is” becomes an equals sign. A word problem is that same translation carried out on a whole situation instead of a single phrase.

Every problem in this lesson can be worked with the same six steps, in this order.

Read the problem and find the question. Decide exactly what quantity you are asked for before doing anything else.
Choose one unknown to name with a variable, and write down what it stands for, units included, as in “let $w$ be the width in centimeters.”
Express every other quantity in the problem in terms of that one variable.
Find the relationship that connects the quantities, and write it as an equation.
Solve the equation.
Check the answer against the words of the original problem, and state the answer to the actual question, with units.

Most of these steps are reading, not algebra. The two that carry the whole method are the third and the sixth. Expressing everything in terms of one variable is what keeps the problem to a single equation you can solve. Checking against the words is what catches a translation that came out backwards. Every problem in this lesson is built so that one quantity can be written in terms of another, which is why one variable is enough for all of them: even when a problem mentions two or three quantities, pin them all to one variable, and a single equation does the job instead of the system you have not learned to solve yet. Not every word problem is built this way; some genuinely need two unknowns and two equations, a tool for a later lesson.

One unknown, everything else in terms of it

The third step is the heart of the method, so it is worth a closer look. A problem often names several quantities, but they are not independent; the words tie them together, and those ties let you write them all with one letter. If a rope is cut into two pieces and one piece is $3$ feet longer than the other, you do not need two variables. Call the short piece $x$ feet, and the long piece is then $x + 3$ feet, because ” $3$ feet longer” means add $3$ . One name, and the relationship supplies the rest.

Consecutive integers are a clean case of this, and they come up often enough to state carefully.

Why consecutive integers are $n$ , $n + 1$ , $n + 2$ #

Consecutive integers, such as $6, 7, 8$ , are defined by a single fact: each one is exactly $1$ more than the one before it.

Name the first of them $n$ . The next integer is ” $1$ more than $n$ ,” which is $n + 1$ , and the one after that is ” $1$ more than $n + 1$ ,” which is $n + 2$ . One unknown, $n$ , names the whole list.

The same idea handles consecutive even integers and consecutive odd integers, with a single change: they step by $2$ instead of $1$ , since one number of the other kind sits between each even number and the next. So consecutive even integers are $n, n + 2, n + 4$ , and consecutive odd integers take the identical form $n, n + 2, n + 4$ . The starting value $n$ has to be even in the first case and odd in the second, but you do not assume which; you let the equation determine $n$ .

The picture behind a “sum” problem is a tape diagram: lay the parts end to end and their combined length is the whole.

Two numbers whose sum is 52. The smaller is a bar of length n and the larger is a bar of length n + 8, so laid end to end they fill the total 52. Writing the larger as n + 8 keeps both numbers tied to the single unknown n.

Worked example 1 The sum of two numbers is $52$ , and the larger is $8$ more than the smaller

The question asks for the two numbers. Name the smaller one: let $n$ be the smaller number. The larger is $8$ more than the smaller, so it is $n + 8$ , and that single relationship pins both numbers to one letter.

The sentence that states an equality is that the sum is $52$ . Translate it:

n + (n + 8) = 52.

Combine like terms and solve:

2n + 8 = 52, \qquad 2n = 44, \qquad n = 22.

So the smaller number is $22$ , and the larger is $n + 8 = 30$ . Check against the words: the sum $22 + 30 = 52$ is right, and $30$ really is $8$ more than $22$ . The two numbers are $22$ and $30$ .

Worked example 2 Find three consecutive integers whose sum is $84$

The question asks for three integers. Name the smallest: let $n$ be the smallest integer. The three consecutive integers are then $n$ , $n + 1$ , and $n + 2$ . The equality in the problem is that they sum to $84$ :

n + (n + 1) + (n + 2) = 84.

Combine and solve:

3n + 3 = 84, \qquad 3n = 81, \qquad n = 27.

The integers are $27$ , $28$ , and $29$ . Checking, $27 + 28 + 29 = 84$ , and they are consecutive, so the answer is $27, 28, 29$ .

Check your understanding

The sum of three consecutive odd integers is $57$ . What is the largest of the three?

17

19

21

23

Age problems

Age problems turn on one idea: everyone ages at the same rate. If you know someone’s age now, then in $k$ years they will be $k$ years older, and $k$ years ago they were $k$ years younger. So name one person’s present age, and use the given relationship to write everyone else’s present age in terms of it. Then shift the whole cast forward or backward together when the problem talks about another time.

Worked example 3 Maria is $4$ times as old as her brother; in $6$ years she will be twice as old as he is

The question asks for their present ages. Let $b$ be the brother’s age now, in years. Maria is $4$ times as old, so she is $4b$ now. In $6$ years each is $6$ years older, so the brother will be $b + 6$ and Maria will be $4b + 6$ . The equality is that at that time Maria’s age is twice her brother’s:

4b + 6 = 2(b + 6).

Distribute and solve:

4b + 6 = 2b + 12, \qquad 2b = 6, \qquad b = 3.

So the brother is $3$ and Maria is $4b = 12$ . Check with the words: now Maria’s $12$ is $4$ times the brother’s $3$ ; in $6$ years she is $18$ and he is $9$ , and $18$ is twice $9$ . Maria is $12$ and her brother is $3$ .

Money problems

Money problems lean on one formula: the total value of a pile of identical coins is the number of coins times the value of each. Ten dimes are worth $10 \times 10 = 100$ cents, and in general $d$ dimes are worth $10d$ cents. A nickel is worth $5$ cents, a dime $10$ cents, and a quarter $25$ cents. To hold everything to one variable, count one kind of coin with the letter and get the other kind from the total. Working in cents is usually cleanest, because it keeps every value a whole number.

Worked example 4 $30$ coins, all nickels and dimes, worth $245$ cents

The question asks how many of each coin. Let $n$ be the number of nickels. There are $30$ coins in all, so the number of dimes is whatever is left, $30 - n$ . Now write the total value in cents: each nickel is worth $5$ cents and each dime $10$ cents, so

5n + 10(30 - n) = 245.

Distribute and solve:

5n + 300 - 10n = 245, \qquad -5n = -55, \qquad n = 11.

There are $11$ nickels and $30 - 11 = 19$ dimes. Check: $11$ nickels are worth $55$ cents and $19$ dimes are worth $190$ cents, and $55 + 190 = 245$ cents, which is $2$ dollars and $45$ cents. So there are $11$ nickels and $19$ dimes.

Check your understanding

A jar of $20$ coins holds only quarters and dimes and is worth $350$ cents. If $q$ is the number of quarters, which equation models the total value?

25q + 10q = 350

25q + 10(20 - q) = 350

10q + 25(20 - q) = 350

q + (20 - q) = 350

Geometry problems

Geometry problems give you a figure whose measurements are linked by a formula you already know, most often a perimeter. That formula is the equality you translate. Name one length, then use the stated relationships to write the others in terms of it. Finally, set the perimeter equal to its given value.

The width is w, so a length that is 3 more than twice the width is 2w + 3. The perimeter, twice the sum of the length and the width, equals 36.

Worked example 5 A rectangle's length is $3$ cm more than twice its width, and its perimeter is $36$ cm

The question asks for the width and the length. Let $w$ be the width in centimeters. The length is $3$ more than twice the width, which is $2w + 3$ . The perimeter of a rectangle is twice the sum of its length and width, so

2\big((2w + 3) + w\big) = 36.

Simplify inside the parentheses, then solve:

2(3w + 3) = 36, \qquad 6w + 6 = 36, \qquad 6w = 30, \qquad w = 5.

The width is $5$ cm and the length is $2(5) + 3 = 13$ cm. Check: the perimeter is $2(13 + 5) = 36$ cm, and $13$ is $3$ more than twice $5$ . The rectangle is $5$ cm by $13$ cm.

Percent problems

Percent problems are the ones you met in pre-algebra, now written as equations. Recall that a percent is a fraction out of $100$ , so $30\%$ of a quantity means $\frac{30}{100}$ of it, which is $0.3$ times it. The word “of” tells you what the percent is taken of, and “is” tells you where the equals sign goes, so a sentence like ” $18$ is $30\%$ of what number” becomes an equation directly.

A discount works the same way in reverse. Removing $20\%$ of a price leaves the other $80\%$ of it, so a $20\%$ discount on a $50$ dollar price leaves $0.80 \times 50 = 40$ dollars. To find what is paid after a discount, multiply by the percent that remains, not the percent that was taken off.

Worked example 6 A student answered $85\%$ of the questions correctly, getting $34$ right

The question asks how many questions were on the test. Let $t$ be the total number of questions. Getting $85\%$ of them right means

0.85\,t = 34.

Divide both sides by $0.85$ :

t = \frac{34}{0.85} = 40.

So the test had $40$ questions. Check: $85\%$ of $40$ is $0.85 \times 40 = 34$ , which matches. A count of questions must be a whole number, and $40$ is one, so the answer makes sense in context: the test had $40$ questions.

Check your understanding

After a $20\%$ discount, a backpack sells for $36$ dollars. What was its original price, in dollars?

43.20

45

30

56

Distance problems

Distance problems rest on one relationship, distance equals rate times time, written $d = rt$ . Pick the one unknown, usually the time, and write each distance as its rate times its time. Then translate the sentence that links the distances, such as “they are $70$ miles apart” or “they meet.”

Worked example 7 Two cyclists leave the same point at the same time, riding in opposite directions at $12$ and $16$ mph; when are they $70$ miles apart?

The question asks for the time. Let $t$ be the number of hours since they started. Using $d = rt$ , in $t$ hours one cyclist covers $12t$ miles and the other covers $16t$ miles. Because they ride in opposite directions, the distance between them is the sum of the two distances, and the problem sets that sum to $70$ :

12t + 16t = 70.

Combine and solve:

28t = 70, \qquad t = \frac{70}{28} = 2.5.

After $2.5$ hours they are $70$ miles apart. Check: in $2.5$ hours one has gone $12 \times 2.5 = 30$ miles and the other $16 \times 2.5 = 40$ miles, and $30 + 40 = 70$ . The time is $2.5$ hours.

Mixture problems (optional)

This last problem type is optional: it applies an idea you already have, so treat it as extra practice rather than new material. A mixture problem is the coin problem again, with pounds or liters standing in for coins: the value of each ingredient is still its amount times its unit value. Name the amount of one ingredient, get the other from the total, and set the combined value equal to the value of the whole batch.

Worked example 8 Mixing peanuts at $3$ dollars a pound with cashews at $8$ dollars a pound

Suppose you want $10$ pounds of a mix worth $5$ dollars a pound. The blend uses peanuts at $3$ dollars a pound and cashews at $8$ dollars a pound, and the question asks how many pounds of each. Let $p$ be the pounds of peanuts; the rest of the $10$ pounds is cashews, so there are $10 - p$ pounds of cashews. The value of the peanuts plus the value of the cashews equals the value of the mix:

3p + 8(10 - p) = 5 \times 10.

Solve:

3p + 80 - 8p = 50, \qquad -5p = -30, \qquad p = 6.

Use $6$ pounds of peanuts and $10 - 6 = 4$ pounds of cashews. Check: the peanuts are worth $3 \times 6 = 18$ dollars and the cashews $8 \times 4 = 32$ dollars, a total of $50$ dollars for $10$ pounds, which is $5$ dollars a pound. So use $6$ pounds of peanuts and $4$ pounds of cashews.

Common mistakes

Answering the wrong question. Solve for the variable, but then report what the problem actually asked for. If $n$ is the smaller number and the question wants the larger, give $n + 8$ , not $n$ .
Translating “less than” and “more than” backwards. ” $5$ less than $x$ ” is $x - 5$ , not $5 - x$ ; the amount you take away comes second. In the same way ” $3$ less than twice a number” is $2n - 3$ .
Using more variables than the problem needs. Whenever the words tie one quantity to another, pin the second to the first instead of naming it separately. Writing the dimes as a new letter instead of $30 - n$ turns a solvable single equation into a system you are not yet set up to solve.
Mixing units. In a coin problem keep everything in cents (or everything in dollars), never a blend of the two. In a rate problem keep hours with hours and miles with miles.
Skipping the check against the words. Substituting your answer into your own equation only confirms the arithmetic, not the translation. Read the answer back into the original sentences, and make sure it makes sense: a count of people or coins cannot be negative or a fraction.

Key takeaways

Each problem in this lesson is solved by translation: name one unknown with its units, write every other quantity in terms of it, turn the key relationship into an equation, solve, and check against the words.
Consecutive integers are $n, n + 1, n + 2$ ; consecutive even or odd integers are $n, n + 2, n + 4$ .
Ages: in $k$ years everyone is $k$ years older, and $k$ years ago everyone was $k$ years younger. Money and mixtures: total value is amount times unit value (work in cents for coins).
Geometry: the perimeter formula is your equation. Percent: “of” multiplies, $p\%$ means $\frac{p}{100}$ , and a discount leaves the percent that remains. Distance: $d = rt$ .
Always check the answer against the original words, report the quantity actually asked for with its units, and confirm it makes sense in context.

Word Problems with Linear Equations

Learning goals

Reading a problem as an equation

One unknown, everything else in terms of it

Age problems

Money problems

Geometry problems

Percent problems

Distance problems

Mixture problems (optional)

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice