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Grade 3 · 3.OA.D.8 · 3.OA.A.4

Using a letter for the unknown number

Eight crayon boxes hold sixty-four crayons in all. How do you write that story as one equation when one of its numbers is not known yet?

Where you are 5 in Variables

When you reach for it

Until now a missing number in an equation has been marked with a box. In Finding the missing number in an equation the box sat in an adding or subtracting fact, such as 8 + \square = 15, and was found by asking what number makes both sides the same. In Equal group problems the box moved into multiplication and division: 6 \times 4 = \square, 8 \times \square = 56, \square \times 5 = 35.

This lesson changes only the mark. In place of the box, write a letter:

8 \times c = 64

The letter does exactly the job the box did. It holds the place of one number that the story has not told us yet. Reach for a letter whenever a story has one number missing and you want to write the whole story down as a single equation before working anything out.

Why change at all, when the box worked? Three reasons, each of which shows up later.

  • A letter can be chosen to remind the reader what it counts: c for the number of crayons, b for the number of baseball cards, m for the number of dollars still missing.
  • A letter can be named in words. “Here c stands for the number of crayons in one box” is a sentence a reader can check against the story. Two boxes in one equation cannot be told apart.
  • Letters are how all later mathematics writes an unknown. The thinking learned here carries straight into equations with more than one step.

The lesson takes three days. Part 1 puts a letter into multiplication and division stories. Part 2 puts a letter into adding and subtracting stories with numbers up to 1{,}000. Part 3 starts from a story alone and writes one equation that models it, with the letter wherever the unknown sits, in all four operations.

The procedure

Every equation in this lesson is a fact with one number hidden. The work has three parts, and each has a reason.

Write one equation that tells the story. Read the story and find its numbers: two are known, one is not. Write the equation the story describes, in the order the story happens, and put a letter where the unknown number sits. Then say in words what the letter stands for.

Find the number from a fact you know. The equation is a true fact with one number covered up. Uncover it with a fact you already know, or with the facts that come with it: every multiplication has two partner divisions and every addition has two partner subtractions (Fact families, Multiplication and division undo each other). For the numbers up to 1{,}000 in adding and subtracting, counting up (Counting up to find a difference) or subtracting (Subtracting three-digit numbers by place value, including across zero) finds it.

Check by putting the number back. Write the number in place of the letter and work out the equation. If both sides agree, the number is the one the story needs. Then answer the story in a sentence, saying what the number counts.

Part 1: a letter in place of the box

Naama has 8 crayon boxes with the same number of crayons in each. There are 64 crayons in all. How many crayons are in one box?

The number of boxes is known (8), the total is known (64), and the number in each box is not. With a box, this was 8 \times \square = 64. With a letter it is

8 \times c = 64

and the sentence that goes with it is: c stands for the number of crayons in one box.

c c c c c c c c 64 crayons 8 boxes
Figure 1Eight crayon boxes with the same number of crayons in each, 64 crayons in all. Each part of the bar is one box, so the equation is 8 \times c = 64.

The bar shows why the letter appears once but means every part. All eight parts hold the same number, because the boxes are equal, so they all carry the same letter c. Eight c’s together make 64.

What number is c? The equation asks: eight times what number is 64? Example 1 answers it from a known fact and checks the answer by putting it back.

Two things about the letter are worth saying plainly, because the whole lesson rests on them.

A letter stands for a number. In this story c is the number of crayons in one box. It is not a short way to write the word “crayons”. The equation 8 \times c = 64 multiplies 8 by a number, and only a number can be multiplied. If a reader is asked “what is c?”, the answer is 8, a number, and the story then says what that number counts.

The same letter means the same number every time it appears in one equation. In 8 \times c = 64 the letter appears once, but in an equation such as c + c = 18 it appears twice, and both times it is the same number. Example 7 works through that.

A letter can be any letter. Most writers pick one that matches the story, but n or k works as well as c. One letter is best avoided in multiplication: the letter x looks very like the times sign \times, so 8 \times x is hard to read. Every equation in this lesson has exactly one letter in it.

The letter can sit in any place, just as the box could. Dividing 45 baseball cards into 5 equal piles and asking how many are in each pile gives 45 \div 5 = b, with the letter on the answer of the division. The partner multiplication, 5 \times b = 45, describes the same piles with the letter on a factor. Both are correct models of the story, and both give the same b.

b b b b b 45 cards in all 5 piles
Figure 2Forty-five baseball cards in 5 equal piles. The number in each pile is unknown, so the equation is 45 \div 5 = b.

Part 2: letters in adding and subtracting stories

Adding and subtracting stories have three numbers too: two parts and the whole they make. A letter can stand for any one of the three.

Yonit keeps her savings in an old pencil box. She has 237 dollars in it and wants 500 dollars for a bicycle. How many more dollars does she need?

The story is an addition with a part missing. What she has, plus what she still needs, makes 500:

237 + m = 500

Here m stands for the number of dollars she still needs. Finding m means finding the number that takes 237 up to 500, and counting up in friendly jumps does it.

237 240 300 500 +3 +60 +200
Figure 3Counting up from 237 to 500 in three jumps, 3 + 60 + 200 = 263. The jumps add up to the missing number in 237 + m = 500.

The partner subtraction, 500 - 237, gives the same number, because adding a part and taking it away undo each other. Example 3 works it through, so m = 263.

The letter could have been in other places in a story with the same three numbers.

The story asks forEquation with a letter
the whole: she has 237 and gets 263 more237 + 263 = m
the part still needed237 + m = 500
the starting amount: some dollars plus 263 make 500m + 263 = 500

A subtraction story works the same way. When something is taken away, the letter can stand for the starting amount, the amount taken, or what is left. Finding it still comes back to one fact, because every subtraction is an addition read the other way: if 443 - 125 = 318, then 318 + 125 = 443.

Part 3: writing one equation for a story

Parts 1 and 2 each started from a story whose equation was easy to see. Part 3 makes that the skill itself: given only the story, write one equation that models it.

An equation models a story when each number in the equation is a number from the story, the operation is what happens in the story, and the letter sits where the story’s unknown number sits. Reading the equation aloud should retell the story.

Two habits make this reliable.

Write the story in the order it happens, letter and all. If the story starts with an unknown amount and then something happens to it, the letter goes first. “Naama shared some crayons equally among 4 friends, and each friend got 7” starts with an unknown number of crayons, which is then divided by 4:

c \div 4 = 7

There is no need to turn the story around before writing it. The equation can be written first and worked out second, which is the whole point of having a letter.

Name the letter in words. “Here c stands for the number of crayons Naama shared.” Saying it catches an equation that has the letter in the wrong place: if the sentence does not match what the story asks, the equation does not model the story.

Then find the number with a fact: which number, divided into 4 equal shares, gives 7 in each? Example 5 finds it with the partner multiplication.

Often more than one equation models the same story, and that is fine. For “512 crayons were in the bin; after some were handed out, 384 were left”, both 512 - h = 384 and 384 + h = 512 are true to the story, since what was left and what was handed out make up the 512. Any equation that models the story gives the same number for the letter.

Worked examples

Examples 1 and 2 go with Part 1, Examples 3 and 4 with Part 2, and Examples 5 to 7 with Part 3.

Example 1

Crayon boxes

Naama has 8 crayon boxes with the same number of crayons in each, 64 crayons in all. Write an equation with a letter, say what the letter stands for, and find how many crayons are in one box.

Solution

The story gives the number of groups (8 boxes) and the total (64 crayons). The size of each group is missing, so the letter goes in the second factor’s place:

8 \times c = 64

Here c stands for the number of crayons in one box.

Eight times what makes 64? The fact 8 \times 8 = 64 answers it, so c = 8. If that fact does not come straight to mind, the partner division asks the same thing: 64 \div 8 = 8.

Check by putting 8 back in place of c: 8 \times 8 = 64, which is the total in the story. Each box holds 8 crayons.

Example 2

Baseball cards in piles

Shira sorts 45 baseball cards into 5 equal piles. Write an equation with a letter for the number of cards in each pile, find it, and check it.

Solution

The cards are shared into 5 equal piles, which is a division, and the number in each pile is the answer to that division:

45 \div 5 = b

Here b stands for the number of cards in one pile.

Division is a multiplication with a factor missing (Division facts), so the question is the same as 5 \times b = 45: five piles of what number make 45? Counting by fives, 5, 10, 15, 20, 25, 30, 35, 40, 45 is nine fives, so 5 \times 9 = 45 and b = 9.

Check by putting 9 back: 45 \div 5 = 9, and multiplying back, 5 \times 9 = 45 cards, all of them. Each pile has 9 cards.

Example 3

Savings in a pencil box

Yonit has 237 dollars saved in her pencil box. A bicycle costs 500 dollars. Write an equation with a letter for the number of dollars she still needs, and find it.

Solution

What she has and what she still needs together make the price. The story happens in that order, so the equation is

237 + m = 500

where m stands for the number of dollars she still needs.

Count up from 237 to 500 in jumps that land on friendly numbers:

237 \xrightarrow{+3} 240 \xrightarrow{+60} 300 \xrightarrow{+200} 500

The jumps make 3 + 60 + 200 = 263, so m = 263. The partner subtraction gives the same number: 500 - 237 = 263.

Check by putting 263 back: 237 + 263 = 500. Yonit needs 263 more dollars.

Example 4

Cards given away

Shira had some baseball cards. She gave 125 of them to Yonit and has 318 left. Write an equation for the story with the letter in the place where the unknown sits, and find how many cards Shira had at the start.

Solution

The unknown is the starting number, and the story then takes 125 away from it, leaving 318. Writing it in that order puts the letter first:

s - 125 = 318

Here s stands for the number of cards Shira had at the start.

Taking away 125 left 318, so putting the 125 back must give the start: 318 + 125 = s. Adding by place, 300 + 100 = 400, 10 + 20 = 30, 8 + 5 = 13, and 400 + 30 + 13 = 443. So s = 443.

Check by putting 443 back: 443 - 125 = 318, the number of cards she has left. Shira started with 443 cards.

Example 5

Shared crayons

Naama shared some crayons equally among 4 friends, and each friend got 7. Write one equation that models the story, say what the letter stands for, and find it.

Solution

The story starts with an unknown number of crayons and divides it into 4 equal shares of 7. Written in the order it happens:

c \div 4 = 7

Here c stands for the number of crayons Naama shared.

The letter is the number being divided, so this is the start of the division that is missing. The partner multiplication finds it: 4 shares of 7 make 4 \times 7 = 28, so c = 28.

Check by putting 28 back: 28 \div 4 = 7. Naama shared 28 crayons.

The same story could also be written 4 \times 7 = c, with the letter on the total. Both model the story, and both give 28.

Example 6

Two equations for one story

There were 512 crayons in the art room bin. Yonit handed some out, and 384 were left. Write two different equations that model the story, find the number of crayons she handed out, and check it in both.

Solution

The crayons handed out and the crayons left together make the 512 that were in the bin.

384 left handed out h 512 crayons
Figure 4The 512 crayons split into the 384 left in the bin and the h handed out. The bar fits both 512 - h = 384 and 384 + h = 512.

Reading the story in the order it happens, 512 crayons, minus the ones handed out, leaves 384:

512 - h = 384

Reading the bar as two parts making a whole gives the second equation:

384 + h = 512

In both, h stands for the number of crayons Yonit handed out.

Count up from 384 to 512: from 384 to 400 is 16, from 400 to 500 is 100, and from 500 to 512 is 12. The jumps make 16 + 100 + 12 = 128, so h = 128.

Check in both equations: 512 - 128 = 384 and 384 + 128 = 512. Both are true, so 128 fits the story either way it is written. Yonit handed out 128 crayons.

Example 7

The same letter twice

Naama puts the same number of crayons into each of 2 crayon boxes, 18 crayons in all. Write an equation using the letter c for the number in one box, and find c. Then explain why the boxes cannot hold 8 and 10, even though 8 + 10 = 18.

Solution

One box holds c crayons and the other box holds c crayons too, because the boxes hold the same number. Together they make 18:

c + c = 18

The same letter appears twice, so it is the same number both times. The question is: which number, added to itself, makes 18? The doubles fact 9 + 9 = 18 answers it, so c = 9.

Two equal boxes are also 2 groups of c, so 2 \times c = 18 models the same story, and 2 \times 9 = 18 gives the same c = 9.

Check by putting 9 back in both places: 9 + 9 = 18. Each box holds 9 crayons.

The pair 8 and 10 does make 18, but putting it into the equation would need c to be 8 in one place and 10 in the other. Inside one equation a letter has one value, so that is not allowed, and the story agrees: boxes with 8 and 10 crayons do not hold the same number.

Boundaries

  • One letter, one number, inside one equation. Every time a letter appears in an equation it stands for the same number. A different equation for a different story may use the same letter for a different number: the c in Example 1 is 8 and the c in Example 5 is 28.
  • A letter stands for a number, not a thing. The letter c means “the number of crayons in one box”, never “crayons”. This is what lets the letter be multiplied, divided, added and subtracted.
  • There may be no whole number that works. In 6 \times r = 40 no whole number fits, because 6 \times 6 = 36 and 6 \times 7 = 42. In 300 + m = 200 no whole number fits either: adding to 300 can never give less than 300. A story that leads to such an equation has been read wrong or has a remainder, which belongs to Remainders and what they mean.
  • There may be more than one number that works. In 0 \times n = 0 every number fits, because zero groups of anything is zero. Every equation in this lesson’s stories has exactly one number that works, but that is a fact about these stories, not about all equations.
  • One step only. Each story here has one action and one unknown number. Stories that need two steps, written with a letter in the same way, are Problems using several operations.
  • Within the facts and within 1{,}000. Multiplying and dividing here stay within the facts up to 10 \times 10 (and a one-digit number times a multiple of ten), and adding and subtracting stay within 1{,}000, so the letter can always be found from a known fact or by counting up.

Where this goes next

The next lesson, Problems using several operations, writes two-step stories with a letter for the unknown, using the same habits: one equation in the order of the story, the letter named in words, and a check by putting the number back. Later, What a letter stands for shows that a letter can also stand for any number at all, as in n + 0 = n, not only for one hidden number. Writing and reading numerical expressions from words turns phrases such as “five more than a number” into expressions, Evaluating expressions and formulas works out an expression once a letter’s number is known, and What a solution is, and checking one and One-step equations give the putting-back check its later name, a solution, and a general way to find it.

Exercises

Part 1: a letter in place of the box

  1. Shira puts 9 baseball cards on each page of her album and fills 7 pages. Write an equation with a letter for the number of cards, and find it.
  2. Yonit has 32 crayons and puts the same number into each of 4 crayon boxes. Write an equation with a letter, say what the letter stands for, and find it.
  3. Naama sorts 54 baseball cards into piles of 6. How many piles does she make? Write an equation with the letter first.
  4. Find the number each letter stands for. (a) 7 \times k = 56 (b) 81 \div 9 = m (c) r \times 4 = 36 (d) 3 \times 30 = t
  5. Write a pencil box story for 6 \times n = 48, say what n stands for, and find n.
  6. Put each number back into its equation to decide whether it makes the equation true. (a) 42 \div g = 7, with g = 6 (b) 9 \times w = 72, with w = 7
  7. A pencil box holds 30 pencils. Yonit’s class has 6 full pencil boxes. Write an equation with a letter for the number of pencils, and find it.

Part 2: letters in adding and subtracting stories

  1. Shira had 146 baseball cards and bought 78 more. Write an equation with a letter for the number she has now, and find it.
  2. Naama had some dollars in her pencil box. She put in 125 more dollars and now has 410 dollars. Write an equation with the letter first, and find how many dollars she had at the start.
  3. Shira has 700 baseball cards and gives 254 of them to Yonit. Write an equation and find how many cards Shira has left.
  4. Yonit’s card album has room for 600 cards, and 428 cards are in it. Write an addition equation with a letter for the number of cards still needed to fill it, and find it.
  5. Find the number each letter stands for, and check it by putting it back. (a) 360 + k = 900 (b) w - 200 = 475 (c) 813 - d = 600
  6. Write a crayon story for 250 - n = 180, say what n stands for, and find n.

Part 3: writing one equation for a story

For each story in 14 to 17, write one equation that models it, say what the letter stands for, find the number, and check it.

  1. Naama shared some crayons equally into 3 crayon boxes, and each box got 9 crayons. How many crayons did she share?
  2. Shira had some baseball cards and sold 160 of them. She has 345 left. How many did she have at the start?
  3. Yonit has 9 pencil boxes holding 54 pencils, the same number in each box. How many pencils are in each box?
  4. Naama has 532 baseball cards. Some are in her album and 248 are in a shoebox. How many are in her album? Write two different equations that model the story.
  5. Which equation models this story? “Shira puts 36 crayons into boxes, 4 in each box. How many boxes does she fill?” (a) 36 \times 4 = b (b) b \times 4 = 36 (c) 4 + b = 36. Find b.
  6. Find the number each letter stands for. (a) n + n = 16 (b) k + k + k = 27
  7. Yonit gives 8 crayons to each of some friends and gives away 48 crayons in all. Write an equation with a letter for the number of friends, and find it.

Explaining and reasoning

  1. Explain why c cannot be 6 in one place and 8 in the other in c + c = 14, even though 6 + 8 = 14. What is c?
  2. In 5 \times c = 40, for a story about crayon boxes, c stands for the number of crayons in one box. Explain why it is not enough to say that c stands for “crayons”.
  3. Explain why 150 + m = 400 and 400 - 150 = m give the same number for m, and find it.
  4. Is there a whole number r that makes 6 \times r = 40 true? Use multiplication facts to decide.
  5. Write a baseball card story that b \div 7 = 6 models, say what b stands for, and find b.
Answers
  1. 7 \times 9 = t, where t is the number of cards. Since 7 \times 9 = 63, t = 63 cards.
  2. 4 \times c = 32, where c is the number of crayons in one box. Since 4 \times 8 = 32, c = 8. Check: 4 \times 8 = 32.
  3. p \times 6 = 54, where p is the number of piles. Since 9 \times 6 = 54, p = 9 piles. As a division, 54 \div 6 = 9.
  4. (a) k = 8, since 7 \times 8 = 56. (b) m = 9, since 9 \times 9 = 81. (c) r = 9, since 9 \times 4 = 36. (d) t = 90, since 3 groups of 3 tens are 9 tens.
  5. One story: Yonit has 48 pencils and puts the same number into each of 6 pencil boxes; how many pencils go in each box? Here n is the number of pencils in one box. Since 6 \times 8 = 48, n = 8. Any story with 6 equal groups making 48 that asks for the size of each group works.
  6. (a) True: 42 \div 6 = 7, because 6 \times 7 = 42. (b) Not true: 9 \times 7 = 63, not 72. The number that works is w = 8, since 9 \times 8 = 72.
  7. 6 \times 30 = p, where p is the number of pencils. Six groups of 3 tens are 18 tens, so p = 180 pencils.
  8. 146 + 78 = t, where t is the number of cards she has now. 146 + 78 = 224, so t = 224.
  9. s + 125 = 410, where s is the number of dollars at the start. 410 - 125 = 285, so s = 285 dollars. Check: 285 + 125 = 410.
  10. 700 - 254 = r, where r is the number of cards left. 700 - 254 = 446, so r = 446 cards. Check: 446 + 254 = 700.
  11. 428 + m = 600, where m is the number of cards still needed. Counting up: 428 to 430 is 2, 430 to 500 is 70, 500 to 600 is 100, so m = 172. Check: 428 + 172 = 600.
  12. (a) k = 540, since 900 - 360 = 540; check 360 + 540 = 900. (b) w = 675, since 475 + 200 = 675; check 675 - 200 = 475. (c) d = 213, since 813 - 600 = 213; check 813 - 213 = 600.
  13. One story: Naama had 250 crayons and gave some to her class; 180 are left. How many did she give? Here n is the number of crayons she gave. Since 180 + 70 = 250, n = 70. Check: 250 - 70 = 180.
  14. c \div 3 = 9, where c is the number of crayons shared. Since 3 \times 9 = 27, c = 27. Check: 27 \div 3 = 9.
  15. s - 160 = 345, where s is the number of cards at the start. Since 345 + 160 = 505, s = 505. Check: 505 - 160 = 345.
  16. 9 \times p = 54, where p is the number of pencils in one box. Since 9 \times 6 = 54, p = 6. Check: 9 \times 6 = 54.
  17. a + 248 = 532 and 532 - 248 = a, where a is the number of cards in the album. 532 - 248 = 284, so a = 284. Check: 284 + 248 = 532.
  18. (b), b \times 4 = 36: b boxes of 4 make 36. Since 9 \times 4 = 36, b = 9 boxes. Equation (a) multiplies the total by the size of a box, and (c) adds where the story has equal groups.
  19. (a) n = 8, since 8 + 8 = 16. (b) k = 9, since 9 + 9 + 9 = 27, which is 3 \times 9 = 27.
  20. f \times 8 = 48, where f is the number of friends. Since 6 \times 8 = 48, f = 6 friends. As a division, 48 \div 8 = 6.
  21. Inside one equation a letter stands for one number, so both places hold the same number. The number that added to itself makes 14 is 7, since 7 + 7 = 14. So c = 7, and 6 and 8 are not allowed because they are two different numbers.
  22. A letter stands for a number. “Crayons” is a thing and cannot be multiplied by 5; “the number of crayons in one box” is a number that can. Saying which number also ties the letter to the story: here c = 8, since 5 \times 8 = 40, and the sentence says that 8 counts the crayons in one box.
  23. Both equations are about the same three numbers: 150 and m are the parts and 400 is the whole. Taking one part from the whole leaves the other part, so the subtraction finds the same missing part the addition hides. 400 - 150 = 250, so m = 250 in both, and 150 + 250 = 400 checks it.
  24. No. 6 \times 6 = 36 and 6 \times 7 = 42, and 40 falls between them, so no whole number of sixes makes exactly 40.
  25. One story: Shira shares her baseball cards equally among 7 friends, and each friend gets 6. How many cards did she share? Here b is the number of cards shared. Since 7 \times 6 = 42, b = 42. Check: 42 \div 7 = 6.
kg1128 · updated 2026-10-21 · CC-BY-SA-4.0