Primary & Middle School Mathematics · Grades 1–9

48Literal Expressions

A letter in a computation stands for a number we do not know yet, or for all numbers at once: writing $P = 4c$ gives the perimeter of every square in one formula. This chapter teaches the writing conventions, how to substitute values, and how to simplify simple expressions — the start of algebra, continued in Chapter 57.

48.1 Writing with letters

Definition 48.1 (Literal expression)

A literal expression is a computation containing one or more letters, each standing for a number. Writing conventions:

the sign $\times$ is dropped before a letter or a bracket: $3 \times a = 3a$ , $a \times b = ab$ , $2 \times (x + 5) = 2(x + 5)$ ;
but it stays between two numbers: $3 \times 5$ cannot be written $35$ !
$a \times a$ is written $a^2$ (“ $a$ squared”), $a \times a \times a = a^3$ ;
$1a$ is written $a$ , and $0a$ is $0$ .

Example 48.2

Simplify the writing:

5 \times x + 3 \times y = 5x + 3y, \qquad a \times 7 = 7a \ \text{(number first)}, \qquad x \times x \times 4 = 4x^2 .

Example 48.3 (Formulas are literal expressions)

For a rectangle of length $L$ and width $w$ : perimeter $P = 2(L + w)$ , area $A = Lw$ . One formula summarizes the infinitely many rectangles at once — that is the whole point of letters.

Two formulas, valid for every rectangle: substitute the values of L and w to get numbers. — Two formulas, valid for every rectangle: substitute the values of $L$ and $w$ to get numbers.

48.2 Substituting

Method 48.4 (Evaluating an expression)

To evaluate an expression for given values of its letters:

rewrite the expression putting the hidden $\times$ signs back: $3a + a^2$ is $3 \times a + a \times a$ ;
replace each letter by its value, keeping brackets around it if needed;
compute, respecting the priorities of Chapter 45.

Example 48.5

Evaluate $E = 3x + x^2$ for $x = 5$ :

E = 3 \times 5 + 5 \times 5 = 15 + 25 = 40 .

Evaluate $F = 2(a + 3b)$ for $a = 4$ and $b = 0.5$ :

F = 2 \times (4 + 3 \times 0.5) = 2 \times (4 + 1.5) = 2 \times 5.5 = 11 .

48.3 Reducing expressions

Proposition 48.6 (Collecting like terms)

Terms containing the same letter part can be collected by adding their number parts (this is distributivity, Theorem 45.6, read backwards):

5x + 3x = (5 + 3)x = 8x, \qquad 7a - 2a = 5a .

Terms with different letter parts (like $3x$ and $5y$ , or $x$ and $x^2$ ) cannot be collected.

Example 48.7

Reduce $E = 4x + 7 + 3x + 2$ , sorting the terms:

E = (4x + 3x) + (7 + 2) = 7x + 9 .

Careful: $7x + 9$ is not $16x$ ! Test with $x = 1$ : $4 + 7 + 3 + 2 = 16$ and $7 \times 1 + 9 = 16$ agree, but $16x$ would also be $16$ — test with $x = 2$ : the original gives $8 + 7 + 6 + 2 = 23$ , and $7 \times 2 + 9 = 23$ , while $16 \times 2 = 32$ . The reduction $7x + 9$ is right, $16x$ is wrong.

Method 48.8 (Testing an equality)

To check whether two expressions might be equal for every value:

substitute the same value in both, at least twice (e.g. $x = 2$ and $x = 10$ ; avoid $0$ and $1$ , they hide too many differences);
if the results differ even once, the expressions are not equal — one counterexample kills a general claim;
if the results agree, the equality is only plausible: a proof needs algebra (expanding, reducing).

Example 48.9

Is $2(x + 3)$ equal to $2x + 6$ for every $x$ ? Expanding proves it: $2(x+3) = 2x + 2 \times 3 = 2x + 6$ . Is $(x + 1)^2$ equal to $x^2 + 1$ ? Test $x = 2$ : $(2+1)^2 = 9$ but $2^2 + 1 = 5$ . No — and one counterexample is a complete proof of “no”.

48.4 Producing expressions

Example 48.10 (From situation to expression)

A taxi charges $4$ euros plus $2$ euros per kilometer. For a trip of $k$ kilometers the price is

p = 4 + 2k \quad\text{euros}.

For $k = 7$ : $p = 4 + 14 = 18$ euros. The expression is the general answer; substituting produces every particular one.

Example 48.11 (Number tricks)

“Think of a number, double it, add $10$ , divide by $2$ , subtract your number.” Call the number $n$ and follow the steps:

n \ \to\ 2n \ \to\ 2n + 10 \ \to\ \frac{2n + 10}{2} = n + 5 \ \to\ n + 5 - n = 5 .

Everyone gets $5$ , whatever $n$ was — the letters explain the magic.

48.5 Exercises

Exercise 48.1 ★

Simplify the writing:

7 \times x, \qquad y \times 3, \qquad a \times b \times 5, \qquad x \times x, \qquad 2 \times (a + 4).

Solution

Solution of Exercise 48.1.

$7x$ ; $3y$ ; $5ab$ ; $x^2$ ; $2(a + 4)$ .

Exercise 48.2 ★

Put the $\times$ signs back:

5ab, \qquad 3(x + 2), \qquad x^2 + 4x, \qquad 2a^3 .

Solution

Solution of Exercise 48.2.

$5ab = 5 \times a \times b$ ; $3(x+2) = 3 \times (x + 2)$ ; $x^2 + 4x = x \times x + 4 \times x$ ; $2a^3 = 2 \times a \times a \times a$ .

Exercise 48.3 ★

Evaluate for $x = 3$ : $5x + 1$ ; $x^2$ ; $2x^2 - x$ ; $10 - 2x$ .

Solution

Solution of Exercise 48.3.

For $x = 3$ : $5 \times 3 + 1 = 16$ ; $3 \times 3 = 9$ ; $2 \times 9 - 3 = 15$ ; $10 - 6 = 4$ .

Exercise 48.4 ★

Evaluate $3a + 2b$ and $a b + 5$ for $a = 4$ , $b = 1.5$ .

Solution

Solution of Exercise 48.4.

$3a + 2b = 12 + 3 = 15$ ; $ab + 5 = 4 \times 1.5 + 5 = 6 + 5 = 11$ .

Exercise 48.5 ★

Reduce:

8x + 5x, \qquad 9a - 4a + a, \qquad 6y + 2 + 3y + 7, \qquad 5t + 3 - 2t - 3 .

Solution

Solution of Exercise 48.5.

$8x + 5x = 13x$ ; $9a - 4a + a = 6a$ ; $6y + 2 + 3y + 7 = 9y + 9$ ; $5t + 3 - 2t - 3 = 3t$ .

Exercise 48.6 ★

Give a formula for: the perimeter of a square of side $c$ ; the perimeter of an isosceles triangle with base $b$ and equal sides $s$ ; the price of $n$ breads at $1.20$ euros each.

Solution

Solution of Exercise 48.6.

Square: $P = 4c$ . Isosceles triangle: $P = b + 2s$ . Breads: $p = 1.20\,n$ euros.

Exercise 48.7 ★

Match each expression to a description: $2n$ ; $n + 2$ ; $n^2$ ; $\frac n2$ . Descriptions: “the half of $n$ ”; “the double of $n$ ”; “the square of $n$ ”; “two more than $n$ ”.

Solution

Solution of Exercise 48.7.

$2n$ : the double; $n + 2$ : two more than $n$ ; $n^2$ : the square; $\frac n2$ : the half.

Exercise 48.8 ★★

Using Method 48.8, decide (with proof or counterexample) whether these equalities hold for every $x$ :

3(x + 4) = 3x + 12; \qquad 2x + 5 = 7x; \qquad x + x = x^2 .

Solution

Solution of Exercise 48.8.

$3(x+4) = 3x + 12$ : true for every $x$ (distributivity — this is a proof, no test needed).

$2x + 5 = 7x$ : test $x = 2$ : left $9$ , right $14$ — false in general (it holds only for $x = 1$ ).

$x + x = x^2$ : test $x = 3$ : left $6$ , right $9$ — false in general ( $x + x = 2x$ ).

Exercise 48.9 ★★

A rectangle has width $w$ and its length is $3$ cm more than its width.

Express the length, then the perimeter, in terms of $w$ , and reduce.
Compute the perimeter for $w = 5$ cm, directly and with your reduced formula. Same answer?

Solution

Solution of Exercise 48.9.

1. Length: $w + 3$ . Perimeter:

P = 2\bigl(w + (w + 3)\bigr) = 2(2w + 3) = 4w + 6 .

2. For $w = 5$ : directly, the sides are $5$ and $8$ , so $P = 2 \times 13 = 26$ cm; with the formula, $4 \times 5 + 6 = 26$ cm. They agree.

Exercise 48.10 ★★

Follow the trick of Example 48.11 with letters: “think of a number, add $6$ , multiply by $3$ , subtract $18$ , divide by your starting number” (assume it is not zero). What does everyone get?

Solution

Solution of Exercise 48.10.

Following the steps with the letter $n$ :

n \ \to\ n + 6 \ \to\ 3(n + 6) = 3n + 18 \ \to\ 3n + 18 - 18 = 3n \ \to\ \frac{3n}{n} = 3 .

Everyone gets $3$ .

Exercise 48.11 ★★★

The figure below is a square of side $a$ with a square corner of side $b$ removed ( $b$ smaller than $a$ ). Write two different expressions for its area (one as a difference, one by cutting the L-shape into two rectangles), and check they agree for $a = 5$ , $b = 2$ .

Solution

Solution of Exercise 48.11.

As a difference: $A = a^2 - b^2$ . By cutting the L into two rectangles: a full-width strip of height $a - b$ and, on top of it, a narrower rectangle:

A = a(a - b) + b(a - b)

(bottom rectangle $a$ wide and $a - b$ tall, top rectangle $a - b$ wide and $b$ tall gives the variant $a(a-b) + (a-b)b$ — the same). For $a = 5$ , $b = 2$ : $a^2 - b^2 = 25 - 4 = 21$ , and $5 \times 3 + 2 \times 3 = 15 + 6 = 21$ . Both expressions agree — they always do: this is the identity $a^2 - b^2 = (a + b)(a - b)$ of Chapter 66 in disguise.

48.6 Problem: The matchstick oracle

Problem 48.1

Weekend problem — growing patterns and their $n$ -th term: three formulas for one pattern, and the famous pattern that breaks at the sixth step

Lay matchsticks in a row of squares: one square, then two, then three … How many matches will the hundredth figure need? Counting them one by one would take all weekend; a formula in $n$ answers instantly for every figure at once — that is exactly what letters are for (Example 48.3). But beware: this problem ends with a celebrated pattern that plays fair for five steps and then betrays everyone who trusted it.

Part I — A row of squares. Figure $1$ is one matchstick square ( $4$ matches); each following figure adds one more square sharing a side with the previous one.

Draw figures $1$ to $4$ and count their matches.
Explain why each new square costs exactly $3$ extra matches, and use this to predict the count for figure $10$ without any formula.
Justify the formula: figure $n$ needs $3n + 1$ matches. (Where does the lonely “ $+1$ ” come from?) Check it against question 1.
Amir reasons differently: “ $4$ matches for the first square, then $3$ for each of the $n - 1$ others”, giving $4 + 3(n - 1)$ . Expand and reduce his expression (Proposition 48.6): does he agree with question 3?
Lena counts a third way: “ $2n$ horizontal matches, and $n + 1$ vertical ones.” Write her formula, reduce it, and conclude. Then answer the opening question: how many matches for figure $100$ ?

Part II — Triangles, tables and going backwards.

A row of matchstick triangles (each new triangle leans on the previous one): count the matches for $1$ , $2$ , $3$ triangles, then justify the formula $2n + 1$ for $n$ triangles.
Going backwards: with exactly $49$ matches, how many triangles can the row have? (Undo the formula step by step.)
Square banquet tables seat one guest per free side. When $n$ tables are pushed into one long row, explain why the seating formula is $2n + 2$ , and check it for $1$ , $2$ , $3$ tables.
A party expects $30$ guests. How many tables does one row need? The caterer instead splits the same tables into two rows of seven: how many guests fit now? Explain the gain with the formula.
Dot pattern: figure $n$ is an L of dots, $n$ dots tall and $n$ dots wide (the corner counted once). Count the dots of figures $1$ , $2$ , $3$ , find the formula, and name the numbers it produces. (Stacked together, these Ls tile a square — a story continued in Problem 57.1.)

Part III — Trust, test, prove.

A classmate claims figure $n$ of Part I needs $3(n + 2) - 5$ matches. Settle the claim by expanding and reducing — no testing needed.
“Squaring makes numbers bigger, so $n^2 \geq n$ for every number $n$ .” Test the claim for $n = 2$ , $3$ , $10$ ; then find a number for which it fails. What did the three successful tests prove, and what does the single failure prove (Method 48.8)?
Mark points on a circle and draw every segment between them, then count the regions inside the circle. Do it for $2$ , $3$ , $4$ and $5$ points (place the points unevenly, so that no three segments cross at one interior point). Record the counts. What formula does the pattern whisper?
Now $6$ points, drawn large and unevenly, counting with great care (there are many small regions). How many regions do you find — and what happened to the conjecture?
The moral, in two sentences: what would have been needed to prove the doubling conjecture, and why were the matchstick formulas of Parts I and II never in such danger? Celebrate with a proved prediction: how many matches for the row of $2\,026$ squares?

Solution

Solution of Problem 48.1.

1. Figures $1$ to $4$ need $4$ , $7$ , $10$ , $13$ matches.

2. A new square shares one side with the previous figure, so only $3$ of its $4$ sides are new matches. From figure $4$ ( $13$ matches), six more squares cost $6 \times 3 = 18$ : figure $10$ needs $13 + 18 = 31$ matches.

3. Reading the row from the left: it starts with one vertical match, and each of the $n$ squares then contributes a top, a bottom and a right side — $3$ matches each. Total: $3n + 1$ ; the “ $+1$ ” is the very first vertical stick. Check: $n = 1, 2, 3, 4$ give $4, 7, 10, 13$ .

4. $4 + 3(n - 1) = 4 + 3n - 3 = 3n + 1$ : Amir’s formula expands to exactly the same reduced expression. Full agreement.

5. Lena: $2n + (n + 1) = 3n + 1$ — the same again. Three ways of seeing, one reduced form. Figure $100$ : $3 \times 100 + 1 = 301$ matches.

6. Counts: $3$ , $5$ , $7$ matches. Each new triangle leans on an existing side and adds only $2$ matches, starting from $3$ : figure $n$ needs $3 + 2(n - 1) = 2n + 1$ matches.

7. Undo $2n + 1 = 49$ : removing the first match leaves $48$ , and each triangle then accounts for $2$ : $48 \div 2 = 24$ triangles.

8. In a row of $n$ tables, each table offers its top and bottom side ( $2n$ seats), and the two end tables offer one extra side each ( $+2$ ): $2n + 2$ seats. Check: $4$ , $6$ , $8$ for $n = 1, 2, 3$ .

9. One row: $2n + 2 \geq 30$ needs $2n \geq 28$ , so $14$ tables. Splitting the same $14$ tables into two rows of $7$ : each row seats $2 \times 7 + 2 = 16$ , together $32$ — two more than the single row, because every new row brings two new ends. (Each split gains $2$ seats; caterers know.)

10. The L of size $n$ has $n$ dots going up and $n$ going across, the corner dot shared: $n + n - 1 = 2n - 1$ dots. Figures $1$ , $2$ , $3$ : $1$ , $3$ , $5$ dots — the odd numbers. Nested one around the next, the Ls fill an $n \times n$ square of dots: the story of Problem 57.1.

11. $3(n + 2) - 5 = 3n + 6 - 5 = 3n + 1$ : the claim reduces to the proved formula, so it is right for every $n$ — established by algebra, not by examples.

12. Tests: $4 \geq 2$ , $9 \geq 3$ , $100 \geq 10$ : all pass. But $n = 0.5$ gives $n^2 = 0.25 < 0.5$ : the claim fails. The three tests proved nothing about all numbers (they only failed to find trouble); the single counterexample proves the general claim false (Method 48.8). A letter stands for every number — decimals included.

13. Regions: $2$ points $\to 2$ ; $3$ points $\to 4$ ; $4$ points $\to 8$ ; $5$ points $\to 16$ . The pattern whispers “each point doubles the count”: $32$ regions expected for $6$ points.

14. A careful count gives $31$ regions — not $32$ . The doubling conjecture is dead: it survived four checks and failed the fifth. (This counterexample is famous enough to have a name, “the circle-regions trap”; no evenly-spaced shortcuts — three chords meeting at one point would merge regions and spoil the count.)

15. A proof would have needed a structural reason why adding a point doubles the regions — and there is none: the doubling was a coincidence of small cases. The matchstick formulas were safe because each was read off the construction itself (each square visibly costs $3$ matches, each table visibly offers $2$ seats): structure, not testing. Proved prediction: $3 \times 2\,026 + 1 = 6\,079$ matches.