Primary & Middle School Mathematics · Grades 1–9

57Literal Calculation and Equations

Grade 7 introduced letters and reduced simple expressions (Chapter 48). This chapter adds the two skills that make algebra powerful: expanding products of brackets, and solving equations — finding the number hidden behind the letter.

57.1 Expanding

Theorem 57.1 (Simple and double distributivity)

For all numbers:

k(a + b) = ka + kb, \qquad (a + b)(c + d) = ac + ad + bc + bd .

Proof. The first rule is Theorem 45.6 (signs included: $k$ and the terms may be negative, Chapter 54). For the second, apply the first rule twice, treating $(c + d)$ as a single number:

(a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd . \qedhere

∎

Example 57.2

Mind the signs at every term:

\begin{align*} 3(2x - 5) &= 6x - 15, \\ -2(4x - 7) &= -8x + 14 && \text{(a minus in front flips both signs)}, \\ (x + 3)(x + 5) &= x^2 + 5x + 3x + 15 = x^2 + 8x + 15, \\ (2x - 1)(x - 4) &= 2x^2 - 8x - x + 4 = 2x^2 - 9x + 4 . \end{align*}

Double distributivity as areas: the rectangle (a+b) × (c+d) splits into four smaller rectangles — one per term of the expansion. — Double distributivity as areas: the rectangle $(a+b) \times (c+d)$ splits into four smaller rectangles — one per term of the expansion.

Method 57.3 (Expand and reduce)

Expand every product, writing each term with its sign;
collect like terms ( $x^2$ with $x^2$ , $x$ with $x$ , numbers with numbers);
check with a value: substitute e.g. $x = 2$ in the original and in the result — equal outcomes catch most mistakes.

57.2 Solving equations

Definition 57.4 (Equation)

An equation is an equality with an unknown number, written $x$ (or any letter). Solving it means finding all the values of $x$ making the equality true — the solutions.

Theorem 57.5 (Balance rules)

An equation keeps exactly the same solutions when:

the same number is added to (or subtracted from) both sides;
both sides are multiplied (or divided) by the same nonzero number.

Proof. Admitted at this level. ∎

An equation is a balanced scale: whatever is done to one pan must be done to the other, and the balance — the equality — survives.

Method 57.6 (Solving a first-degree equation)

Expand and reduce each side if needed;
add or subtract on both sides to gather the $x$ -terms on one side, the plain numbers on the other;
reduce, then divide both sides by the coefficient of $x$ ;
check by substituting the value found into the original equation.

Example 57.7

Solve $7x - 5 = 4x + 16$ :

\begin{align*} 7x - 5 &= 4x + 16 \\ 7x - 4x &= 16 + 5 && \text{(add $5$, subtract $4x$, on both sides)} \\ 3x &= 21 \\ x &= 7 && \text{(divide both sides by 3).} \end{align*}

Check: left side $7 \times 7 - 5 = 44$ ; right side $4 \times 7 + 16 = 44$ . The solution is $7$ .

Example 57.8 (With brackets and negatives)

Solve $3(x - 4) = 5 - 2(x + 1)$ :

\begin{align*} 3x - 12 &= 5 - 2x - 2 && \text{(expand both sides)} \\ 3x - 12 &= 3 - 2x && \text{(reduce)} \\ 5x &= 15 && \text{(add $2x + 12$ to both sides)} \\ x &= 3 . \end{align*}

Check: $3(3 - 4) = -3$ and $5 - 2(3+1) = 5 - 8 = -3$ . Solution: $3$ .

Method 57.9 (Solving a problem with an equation)

Choose the unknown: “let $x$ be …” (the quantity asked for, or a simpler related one);
translate the story into an equation;
solve it;
interpret: answer the original question in a sentence, with units, and check the answer against the story.

Example 57.10

Lucas is $4$ years older than his sister; together their ages sum to $26$ . Let $x$ be the sister’s age; then Lucas is $x + 4$ , and

x + (x + 4) = 26 \ \Longrightarrow\ 2x + 4 = 26 \ \Longrightarrow\ 2x = 22 \ \Longrightarrow\ x = 11 .

The sister is $11$ , Lucas is $15$ . Check: $11 + 15 = 26$ and $15 - 11 = 4$ .

57.3 Order and inequalities

Proposition 57.11 (Operations and order)

Adding the same number to both sides of an inequality preserves it; multiplying both sides by a positive number preserves it; multiplying by a negative number reverses it:

2 < 5 \ \text{ but }\ (-1) \times 2 > (-1) \times 5 .

Proof. Admitted at this level. ∎

Example 57.12

Solve $5 - 2x \leq 11$ : subtract $5$ on both sides, $-2x \leq 6$ ; divide by $-2$ and reverse: $x \geq -3$ . All the numbers greater than or equal to $-3$ are solutions — an equation usually has one solution, an inequality has a whole half-line of them.

57.4 Exercises

Exercise 57.1 ★

Expand and reduce:

5(2x + 3), \qquad -3(4x - 1), \qquad 2(3x - 2) + 4(x + 5).

Solution

Solution of Exercise 57.1.

$5(2x + 3) = 10x + 15$ .

$-3(4x - 1) = -12x + 3$ .

$2(3x - 2) + 4(x + 5) = 6x - 4 + 4x + 20 = 10x + 16$ .

Exercise 57.2 ★

Expand and reduce:

(x + 2)(x + 7), \qquad (3x + 1)(2x - 5), \qquad (x - 4)(x - 6).

Solution

Solution of Exercise 57.2.

$(x+2)(x+7) = x^2 + 7x + 2x + 14 = x^2 + 9x + 14$ .

$(3x+1)(2x-5) = 6x^2 - 15x + 2x - 5 = 6x^2 - 13x - 5$ .

$(x-4)(x-6) = x^2 - 6x - 4x + 24 = x^2 - 10x + 24$ .

Exercise 57.3 ★

Is $x = 3$ a solution of $4x - 5 = 7$ ? Of $2x + 1 = x + 4$ ? Of $x^2 = 6x - 9$ ? (Substitute and compare both sides.)

Solution

Solution of Exercise 57.3.

$4 \times 3 - 5 = 7$ : yes. $2 \times 3 + 1 = 7$ and $3 + 4 = 7$ : yes. $3^2 = 9$ and $6 \times 3 - 9 = 9$ : yes — $x = 3$ solves all three.

Exercise 57.4 ★

Solve, with a check:

x + 9 = 4, \qquad 5x = -35, \qquad \frac{x}{3} = 8, \qquad 2x - 7 = 13 .

Solution

Solution of Exercise 57.4.

$x + 9 = 4$ : $x = -5$ (check: $-5 + 9 = 4$ ).

$5x = -35$ : $x = -7$ .

$\frac x3 = 8$ : $x = 24$ .

$2x - 7 = 13$ : $2x = 20$ , $x = 10$ (check: $20 - 7 = 13$ ).

Exercise 57.5 ★

Solve:

6x + 5 = 2x + 25, \qquad 9 - 4x = x - 11, \qquad 3(x + 2) = 2x + 11 .

Solution

Solution of Exercise 57.5.

$6x + 5 = 2x + 25$ : $4x = 20$ , so $x = 5$ .

$9 - 4x = x - 11$ : $20 = 5x$ , so $x = 4$ .

$3(x + 2) = 2x + 11$ : $3x + 6 = 2x + 11$ , so $x = 5$ .

Exercise 57.6 ★

Solve $4(2x - 3) = 5(x + 6)$ , expanding first, and check your solution.

Solution

Solution of Exercise 57.6.

$8x - 12 = 5x + 30$ , so $3x = 42$ and $x = 14$ . Check: $4(28 - 3) = 100$ and $5(14 + 6) = 100$ .

Exercise 57.7 ★

Translate and solve: “three times a number, decreased by $8$ , equals $19$ ”; “the double of a number increased by $5$ equals the number increased by $12$ ”.

Solution

Solution of Exercise 57.7.

“ $3x - 8 = 19$ ”: $3x = 27$ , $x = 9$ .

“ $2x + 5 = x + 12$ ”: $x = 7$ .

Exercise 57.8 ★★

A rectangle’s length is $5$ cm more than its width, and its perimeter is $58$ cm. Let $x$ be the width: write an equation, solve it and give the two dimensions.

Solution

Solution of Exercise 57.8.

Width $x$ , length $x + 5$ ; perimeter $2\bigl(x + (x + 5)\bigr) = 4x + 10 = 58$ , so $4x = 48$ and $x = 12$ : width $12$ cm, length $17$ cm (check: $2 \times (12 + 17) = 58$ ).

Exercise 57.9 ★★

Three consecutive whole numbers have sum $141$ . Call the middle one $n$ and find all three.

Solution

Solution of Exercise 57.9.

The numbers are $n - 1$ , $n$ , $n + 1$ with sum $3n = 141$ , so $n = 47$ : the numbers are $46$ , $47$ , $48$ .

Exercise 57.10 ★★

Solve the inequalities and describe the solutions:

3x + 4 > 19, \qquad 7 - 2x \geq 1, \qquad 5(x - 1) < 2x + 7 .

Solution

Solution of Exercise 57.10.

$3x + 4 > 19$ : $3x > 15$ , so $x > 5$ .

$7 - 2x \geq 1$ : $-2x \geq -6$ , dividing by $-2$ reverses: $x \leq 3$ .

$5(x-1) < 2x + 7$ : $5x - 5 < 2x + 7$ , so $3x < 12$ and $x < 4$ .

Exercise 57.11 ★★

Mia has $140$ euros in savings and adds $15$ euros each month; Tom has $200$ euros and adds $10$ euros each month. After how many months will Mia have strictly more than Tom? (Set up an inequality.)

Solution

Solution of Exercise 57.11.

After $n$ months, Mia has $140 + 15n$ and Tom $200 + 10n$ . Mia leads when

140 + 15n > 200 + 10n \ \Longleftrightarrow\ 5n > 60 \ \Longleftrightarrow\ n > 12 :

from the $13$ th month on.

Exercise 57.12 ★★★

Solve the equation $\dfrac{x + 3}{4} = \dfrac{2x - 1}{5}$ . (Multiply both sides by $20$ , the common denominator, then proceed as usual; check your solution in the original equation.)

Solution

Solution of Exercise 57.12.

Multiply both sides by $20$ :

5(x + 3) = 4(2x - 1) \ \Longrightarrow\ 5x + 15 = 8x - 4 \ \Longrightarrow\ 19 = 3x \ \Longrightarrow\ x = \frac{19}{3}.

Check: $\frac{19/3 + 3}{4} = \frac{28/3}{4} = \frac{28}{12} = \frac73$ , and $\frac{2 \times 19/3 - 1}{5} = \frac{35/3}{5} = \frac{35}{15} = \frac73$ . Equal: the solution is $\frac{19}{3}$ .

57.5 Problem: Remarkable identities and the staircase of odd numbers

Problem 57.1

Weekend problem — $(a+b)^2$ , $(a-b)^2$ , $(a+b)(a-b)$ , and the sum $1 + 3 + 5 + \dots + (2n-1) = n^2$

Three expansions come up so often that algebra knows them by heart: they are called the remarkable identities. This problem derives them from double distributivity (Theorem 57.1), turns them into a mental-arithmetic superpower, uses them to prove a beautiful fact about odd numbers, and ends by solving equations of a brand-new kind: equations with a square in them.

Part I — The three identities.

Expand $(a + b)^2 = (a + b)(a + b)$ and reduce, to prove

$(a + b)^2 = a^2 + 2ab + b^2 .$
Prove likewise the two companions:

$(a - b)^2 = a^2 - 2ab + b^2, \qquad (a + b)(a - b) = a^2 - b^2 .$
Draw a square of side $a + b$ , split each side into $a$ and $b$ , and cut the square into four pieces. Which piece of the first identity does each part of the picture represent?
Mental arithmetic: use the identities to compute, without posing any multiplication,

$101^2, \qquad 99^2, \qquad 49 \times 51 .$
A classic trap: Zoe writes “ $(a + b)^2 = a^2 + b^2$ ”. Test her formula with $a = 3$ , $b = 4$ , and explain on the picture of question 3 exactly which pieces her formula forgets.

Part II — The staircase of odd numbers.

Compute $1$ , then $1 + 3$ , then $1 + 3 + 5$ , then $1 + 3 + 5 + 7$ , then $1 + 3 + 5 + 7 + 9$ . What do you notice about the results?
Explain why the $n$ -th odd number is $2n - 1$ (check: $n = 1, 2, 3$ give $1, 3, 5$ ).
Use an identity of Part I to expand and reduce $(n + 1)^2 - n^2$ . What does the result have to do with question 7?
Question 8 says: to pass from a square of side $n$ to a square of side $n + 1$ , one adds exactly the next odd number of unit cells — an L-shaped border along two sides and the corner. Starting from $1 = 1^2$ and growing the square one layer at a time, explain why

$1 + 3 + 5 + \dots + (2n - 1) = n^2 :$

the sum of the first $n$ odd numbers is the $n$ -th square.
Deduce, without adding them up one by one: the value of $1 + 3 + 5 + \dots + 99$ ; and how many odd numbers, starting from $1$ , add up to exactly $400$ .

Part III — Equations with a square.

Recognize a remarkable identity in $x^2 + 6x + 9$ , and solve the equation $x^2 + 6x + 9 = 0$ .
Explain, using the sign-and-distance rule of Theorem 54.1, why a product of two numbers can be zero only if one of the two factors is zero. Use this to solve $(x - 2)(x + 5) = 0$ .
Solve $x^2 = 49$ by moving everything to one side and factoring $x^2 - 49$ with the third identity. How many solutions does this equation have — and how does that differ from every equation solved in this chapter so far?
In Exercise 57.3 you checked that $x = 3$ is a solution of $x^2 = 6x - 9$ . Show that it is the only one: bring everything to one side, recognize an identity, and conclude.
Finale: compute $2026^2 - 2025^2$ in your head, and explain with an identity why the difference of two consecutive squares is always the sum of the two numbers — which is question 8 all over again.

Solution

Solution of Problem 57.1.

1. Double distributivity (Theorem 57.1) with $c = a$ , $d = b$ :

(a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2 ,

since $ab$ and $ba$ are the same product.

2. Same expansion, minding the signs:

(a - b)(a - b) = a^2 - ab - ba + b^2 = a^2 - 2ab + b^2 ,

(a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - b^2 :

this time the cross-terms $-ab$ and $+ba$ cancel instead of doubling.

3. The square of side $a + b$ cuts into four pieces: a square $a \times a$ (the $a^2$ ), a square $b \times b$ (the $b^2$ ), and two rectangles $a \times b$ (together, the $2ab$ ). The identity $(a+b)^2 = a^2 + 2ab + b^2$ is the statement that the four areas fill the big square — it is the picture of double distributivity with both sides split the same way.

4. With the identities:

\begin{align*} 101^2 &= (100 + 1)^2 = 10\,000 + 2 \times 100 + 1 = 10\,201, \\ 99^2 &= (100 - 1)^2 = 10\,000 - 200 + 1 = 9\,801, \\ 49 \times 51 &= (50 - 1)(50 + 1) = 50^2 - 1^2 = 2\,499 . \end{align*}

5. For $a = 3$ , $b = 4$ : $(3 + 4)^2 = 49$ but $3^2 + 4^2 = 9 + 16 = 25$ . Zoe’s formula forgets the two $a \times b$ rectangles of the picture — the missing $2ab = 24$ , and indeed $25 + 24 = 49$ .

6. $1$ ; $1 + 3 = 4$ ; $1 + 3 + 5 = 9$ ; $1 + 3 + 5 + 7 = 16$ ; $1 + 3 + 5 + 7 + 9 = 25$ . The results are the perfect squares $1^2, 2^2, 3^2, 4^2, 5^2$ .

7. The odd numbers start at $1$ and go up in steps of $2$ : the $n$ -th one is reached after $n - 1$ steps of $2$ from $1$ , so it equals $1 + 2(n - 1) = 2n - 1$ . Check: $n = 1, 2, 3$ give $1, 3, 5$ .

8. By the first identity,

(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 :

the difference between two consecutive squares is exactly the $(n+1)$ -th odd number (question 7 with $n + 1$ in place of $n$ : $2(n+1) - 1 = 2n + 1$ ).

9. Build the squares one after the other. Start with a single cell: $1 = 1^2$ . To pass from the square of side $1$ to the square of side $2$ , add $2 \times 1 + 1 = 3$ cells (question 8): $1 + 3 = 2^2$ . To pass to side $3$ , add $2 \times 2 + 1 = 5$ cells: $1 + 3 + 5 = 3^2$ . Each new odd number is exactly the L-shaped layer completing the next square, so after adding the first $n$ odd numbers the square of side $n$ is complete: $1 + 3 + \dots + (2n - 1) = n^2$ .

10. The odd numbers from $1$ to $99$ are the first $50$ of them ( $2n - 1 = 99$ gives $n = 50$ ), so

1 + 3 + 5 + \dots + 99 = 50^2 = 2\,500 .

And $1 + 3 + \dots + (2n - 1) = 400 = 20^2$ requires $n = 20$ : the first twenty odd numbers ( $1$ up to $39$ ) add up to $400$ .

11. $x^2 + 6x + 9 = x^2 + 2 \times 3 \times x + 3^2 = (x + 3)^2$ : the first identity with $a = x$ , $b = 3$ . So the equation reads $(x + 3)^2 = 0$ . A square is zero only when the number itself is zero (Theorem 54.1: a nonzero number has nonzero distance, so its square has nonzero distance too). Hence $x + 3 = 0$ : the only solution is $x = -3$ .

12. By Theorem 54.1, the distance of a product is the product of the distances. If both factors are nonzero, both distances are nonzero, so the product’s distance is nonzero: the product cannot be $0$ . A zero product therefore forces at least one zero factor. For $(x - 2)(x + 5) = 0$ : either $x - 2 = 0$ or $x + 5 = 0$ , giving the two solutions $x = 2$ and $x = -5$ .

13. $x^2 = 49$ becomes $x^2 - 49 = 0$ , and the third identity factors it:

x^2 - 49 = x^2 - 7^2 = (x - 7)(x + 7) = 0 .

By question 12, $x = 7$ or $x = -7$ : two solutions. Every equation solved so far in this chapter was of first degree and had exactly one solution; a square in the equation can produce two.

14. Bring everything to the left side: $x^2 - 6x + 9 = 0$ , and recognize the second identity: $x^2 - 2 \times 3 \times x + 3^2 = (x - 3)^2$ . So the equation is $(x - 3)^2 = 0$ , which forces $x - 3 = 0$ (as in question 11): $x = 3$ is the one and only solution — the check of Exercise 57.3 found it, and the factoring proves there are no others.

15. By the third identity,

2026^2 - 2025^2 = (2026 - 2025)(2026 + 2025) = 1 \times 4051 = 4\,051 .

In general $(n+1)^2 - n^2 = (n + 1 - n)(n + 1 + n) = 2n + 1$ : the difference of two consecutive squares is the sum of the two numbers — the same $2n + 1$ as in question 8, now obtained from the third identity instead of the first.