Turning Word Problems into Multi-Step Equations | Grade 8 Math | LevelKite

A word problem tells a short story about amounts: a cost, a time, a number of things. One amount is unknown, and somewhere in the story the same amount is described in two ways.

You give the unknown a letter, called a variable. You write each description as an expression, which is a combination of numbers, variables and operations with no equals sign. Then you join the two expressions with an equals sign. The result is an equation: a statement that two expressions have the same value. A solution is a value of the variable that makes the equation true.

You already know how to solve multi-step equations. The new skill is building the right equation from the words and deciding what its solution means in the story.

Here is a short story to translate. A bike shop charges 6 dollars per hour plus a helmet fee of 9 dollars, paid once. Sam pays 33 dollars. For how many hours did Sam rent the bike?

The cost of Sam's rental is described twice: once by the shop's prices and once by what Sam paid. Before you write anything, name the unknown, its unit and its domain, which is the set of values the story permits. The table translates the story one piece at a time.

Translating the bike story into symbols

What the story says	What you write	Why
Sam rents for an unknown number of hours.	$h$ = hours rented, with $h \ge 0$	The unknown gets a letter, a unit and a domain. Time cannot be negative.
The shop charges 6 dollars per hour.	$6h$ dollars	Each of the $h$ hours costs 6 dollars.
The helmet fee is 9 dollars, paid once.	$6h + 9$ dollars	A one-time fee is added once. It is not multiplied by $h$ .
Sam pays 33 dollars.	$6h + 9 = 33$	Both sides name the same cost, in dollars.

Now solve with steps you know. Subtract $9$ from both sides, which keeps the two sides equal: $6h = 24$ . Divide both sides by $6$ , which is allowed because $6$ is not zero: $h = 4$ .

Check in the original equation: $6(4) + 9 = 24 + 9 = 33$ , which is true. The value $4$ is in the domain, so Sam rented the bike for $4$ hours.

Look at the order of the work: you defined the variable, wrote the equation, solved it and checked it. Each worked example in this lesson follows the same four moves.

Key concept

Same amount, two expressions

An equation models a story when both sides name the same amount in the same unit.

Define the variable: what it stands for, its unit and its domain.
Write two expressions for the same amount and set them equal.
Solve. You may replace a side with an equivalent expression (one with the same value for every permitted value of the variable), add or subtract the same quantity on both sides, or multiply or divide both sides by the same nonzero number. These moves keep the solutions the same. Never divide by zero.
Check the result in the original equation and in the story, and confirm it is in the domain. Then answer the question in a sentence.

Worked example

A total for a group

Four friends go to a climbing gym. Each friend pays the same entry fee and also pays 3 dollars to rent shoes. Together they pay 68 dollars. What is the entry fee for one person?

Define the variable. Let f be the entry fee for one person, in dollars. A fee cannot be negative, so f is 0 or more.
$f \ge 0$
Write the equation. One friend pays f + 3 dollars. All four friends pay that same amount, so the group pays 4 times the whole amount f + 3. Parentheses keep f + 3 together. The story says the group pays 68 dollars, and both sides are in dollars.
$4(f + 3) = 68$
Expand the left side with the distributive property: multiply the 4 by each term inside the parentheses. A term is a part that is added or subtracted. So 4 times f is 4f, and 4 times 3 is 12. The two forms are equal for every value of f, so the solutions do not change.
$4f + 12 = 68$
Subtract 12 from both sides. Subtracting the same number from both sides keeps them equal, and 68 - 12 = 56.
$4f = 56$
Divide both sides by 4. Dividing both sides by the same nonzero number keeps the solutions the same, and 4 is not zero. Then 56 divided by 4 is 14.
$f = 14$
Check in the original equation and in the story. One friend pays 14 + 3 = 17 dollars, and four friends pay 4 times 17, which is 68 dollars. The value 14 is in the domain.
$4(14 + 3) = 4 \cdot 17 = 68$

Answer

The entry fee for one person is $14$ dollars.

Worked example

A fraction of the whole

Lena has read $\frac{1}{3}$ of the pages in her book. After she reads 40 more pages, 80 pages will be left. How many pages are in the book?

Define the variable. Let p be the number of pages in the book. Pages are counted, so p is a positive whole number.
Write the equation. The whole book is made of three parts: the pages already read, which is one third of p, then the 40 pages Lena reads next, then the 80 pages left. The parts add up to the whole book, p. Both sides count pages.
$\frac{1}{3}p + 40 + 80 = p$
Collect like terms on the left. Like terms have the same variable part, and 40 and 80 are like terms because neither has a variable. Since 40 + 80 = 120, the left side keeps the same value for every p, so the solutions do not change.
$\frac{1}{3}p + 120 = p$
Subtract one third of p from both sides to bring the p terms together. Subtracting the same quantity from both sides keeps them equal. On the right, p is three thirds of p, and three thirds minus one third is two thirds.
$120 = \frac{2}{3}p$
Multiply both sides by three halves. Three halves is the reciprocal of two thirds, which means their product is 1, so p stands alone on the right. Multiplying both sides by the same nonzero number keeps the solutions the same.
$p = 120 \cdot \frac{3}{2} = \frac{360}{2} = 180$
Check in the original equation and in the story. One third of 180 is 60 pages already read. Then 60 + 40 + 80 = 180, which matches the whole book, and 180 is a positive whole number.
$\frac{1}{3}(180) + 40 + 80 = 60 + 40 + 80 = 180$

Answer

Lena's book has $180$ pages.

Worked example

Two amounts that become equal

Ravi has 15 dollars and saves 7.50 dollars each week. Mina has no savings. She waits 2 weeks and then saves 10 dollars each week. How many weeks from now will they have the same amount, and how much will each of them have?

Define the variable and write the equation. Let w be the number of weeks from now. Savings are added once a week, so w is a whole number, and w must be at least 2 because Mina's saving has not started before then. Ravi has 15 + 7.5w dollars. Mina saves for w - 2 weeks, so she has 10(w - 2) dollars. Set the two amounts equal; both sides are in dollars.
$15 + 7.5w = 10(w - 2)$
Expand the right side with the distributive property: 10 times w is 10w, and 10 times 2 is 20. The 20 is subtracted because the 2 is subtracted inside the parentheses. The two forms are equal for every value of w, so the solutions do not change.
$15 + 7.5w = 10w - 20$
Subtract 7.5w from both sides, which keeps the two sides equal. On the right, the like terms combine: 10w - 7.5w = 2.5w.
$15 = 2.5w - 20$
Add 20 to both sides, which keeps the two sides equal, and 15 + 20 = 35.
$35 = 2.5w$
Divide both sides by 2.5. It is not zero, so the solutions stay the same. The result is 14 because 2.5 times 14 is 35.
$w = 35 \div 2.5 = 14$
Check in the original equation and in the story. Ravi has 15 + 105 = 120 dollars, since 7.5 times 14 is 105. Mina has saved for 14 - 2 = 12 weeks, so she has 10 times 12 = 120 dollars. The amounts match, and 14 is a whole number that is at least 2.
$15 + 7.5(14) = 15 + 105 = 120 \quad \text{and} \quad 10(14 - 2) = 10 \cdot 12 = 120$

Answer

After $14$ weeks, Ravi and Mina will each have exactly $120$ dollars.

In the savings story, exactly one value of $w$ made the two amounts equal. That does not always happen. Compare three ride passes for $r$ rides, where $r$ is a whole number that is $0$ or more.

Pass A is a 10-dollar card plus 2 dollars per ride: $10 + 2r$ dollars.
Pass B is a 15-dollar card plus 2 dollars per ride: $15 + 2r$ dollars.
Pass C charges 2 dollars per ride and bills you for 5 extra rides: $2(r + 5)$ dollars.

A and B: $10 + 2r = 15 + 2r$ . Subtract $2r$ from both sides: $10 = 15$ . This is false, so the equation is a contradiction: no value of $r$ is a solution, and the two passes never cost the same.

A and C: $10 + 2r = 2(r + 5)$ . Expand the right side: $10 + 2r = 2r + 10$ . Subtract $2r$ from both sides: $10 = 10$ . This is true, so the equation is an identity: every permitted value of $r$ is a solution, and the two passes always cost the same.

Takeaway. A word problem becomes an equation when you find one amount described in two ways. Define the variable with its unit and domain, write the two expressions and set them equal, solve with moves that keep the solutions the same, and check the result in the original equation and in the story. If the variable drops out, a true statement means every permitted value works, and a false statement means no value works.