---
title: "Literal Calculation and Equations"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 57
exercises: 12
source: https://one-course.com/books/math/1/en/chapter/57-literal-calculation-and-equations
---

# Chapter 57 — Literal Calculation and Equations

Grade 7 introduced letters and reduced simple expressions ([Chapter 48](https://one-course.com/books/math/1/en/chapter/48-literal-expressions#ch-g7-literal)). This chapter adds the two skills that make algebra powerful: *expanding* [products](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of brackets, and *solving [equations](#def-g8-equations-def)* — 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](https://one-course.com/books/math/1/en/chapter/45-priorities-of-operations#thm-g7-priorities-distributivity) (signs included: $k$ and the terms may be negative, [Chapter 54](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#ch-g8-negprod)). 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.](https://one-course.com/images/onecourse/chapters/math-1/g8-equations/fig-5157e40e0e48.svg)

*Double distributivity as [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area): the rectangle $(a+b) \times (c+d)$ splits into four smaller rectangles — one per term of the expansion.*

**Method 57.3 (Expand and reduce).**

1. Expand every [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) , writing each term with its sign;
2. collect like terms ( $x^2$ with $x^2$ , $x$ with $x$ , numbers with numbers);
3. 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](#def-g8-equations-def) keeps exactly the same solutions when:

1. the same number is added to (or subtracted from) both sides;
2. 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.](https://one-course.com/images/onecourse/chapters/math-1/g8-equations/fig-32043be11d88.svg)

*An [equation](#def-g8-equations-def) 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).**

1. Expand and reduce each side if needed;
2. add or subtract on both sides to gather the $x$ -terms on one side, the plain numbers on the other;
3. reduce, then divide both sides by the coefficient of $x$ ;
4. *check* by substituting the value found into the original [equation](#def-g8-equations-def) .

**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).**

1. Choose the unknown: “let $x$ be …” (the quantity asked for, or a simpler related one);
2. translate the story into an [equation](#def-g8-equations-def) ;
3. solve it;
4. 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](#def-g8-equations-def) 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 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 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 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 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 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 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 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](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) is $58$ cm. Let $x$ be the width: write an [equation](#def-g8-equations-def), solve it and give the two dimensions.

**Solution of Exercise 57.8.**

Width $x$, length $x + 5$; [perimeter](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $141$. Call the middle one $n$ and find all three.

**Solution of Exercise 57.9.**

The numbers are $n - 1$, $n$, $n + 1$ with [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $3n = 141$, 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 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 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](#def-g8-equations-def) $\dfrac{x + 3}{4} = \dfrac{2x - 1}{5}$. (Multiply both sides by $20$, the common [denominator](https://one-course.com/books/math/1/en/chapter/24-first-fractions#def-g4-fractions-def), then proceed as usual; check your solution in the original [equation](#def-g8-equations-def).)

**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](#thm-g8-equations-expand)), turns them into a mental-arithmetic superpower, uses them to prove a beautiful fact about [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), and ends by solving [equations](#def-g8-equations-def) of a brand-new kind: [equations](#def-g8-equations-def) with a square in them.

**Part I — The three identities.**

1. Expand $(a + b)^2 = (a + b)(a + b)$ and reduce, to prove $$(a + b)^2 = a^2 + 2ab + b^2 .$$
2. Prove likewise the two companions: $$(a - b)^2 = a^2 - 2ab + b^2,  \qquad  (a + b)(a - b) = a^2 - b^2 .$$
3. 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?
4. Mental arithmetic: use the identities to compute, without posing any multiplication, $$101^2, \qquad 99^2, \qquad 49 \times 51 .$$
5. 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd).**

6. 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?
7. Explain why the $n$ -th [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is $2n - 1$ (check: $n = 1, 2, 3$ give $1, 3, 5$ ).
8. 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?
9. 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) 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](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the first $n$ [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is the $n$-th square.
10. Deduce, without adding them up one by one: the value of $1 + 3 + 5 + \dots + 99$ ; and how many [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) , starting from $1$ , add up to exactly $400$ .

**Part III — [Equations](#def-g8-equations-def) with a square.**

11. Recognize a remarkable identity in $x^2 + 6x + 9$ , and solve the [equation](#def-g8-equations-def) $x^2 + 6x + 9 = 0$ .
12. Explain, using the sign-and-distance rule of [Theorem 54.1](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#thm-g8-negprod-rules) , why a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of two numbers can be [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) *only if* one of the two factors is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) . Use this to solve $(x - 2)(x + 5) = 0$ .
13. 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](#def-g8-equations-def) have — and how does that differ from every [equation](#def-g8-equations-def) solved in this chapter so far?
14. In [Exercise 57.3](#exo-g8-equations-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.
15. Finale: compute $2026^2 - 2025^2$ in your head, and explain with an identity why the [difference](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of two consecutive squares is always the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two numbers — which is question 8 all over again.

**Solution of Problem 57.1.**

**1.** Double distributivity ([Theorem 57.1](#thm-g8-equations-expand)) 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](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def).

**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](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) 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](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) between two consecutive squares is exactly the $(n+1)$-th [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) (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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) is exactly the L-shaped layer completing the next square, so after adding the first $n$ [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) the square of side $n$ is complete: $1 + 3 + \dots + (2n - 1) = n^2$.

**10.** The [odd numbers](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) 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](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) ($1$ 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](#def-g8-equations-def) reads $(x + 3)^2 = 0$. A square is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) only when the number itself is [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) ([Theorem 54.1](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#thm-g8-negprod-rules): 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](https://one-course.com/books/math/1/en/chapter/54-multiplying-negative-numbers#thm-g8-negprod-rules), the distance of a [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) is the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) of the distances. If both factors are nonzero, both distances are nonzero, so the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def)’s distance is nonzero: the [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) cannot be $0$. A [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) [product](https://one-course.com/books/math/1/en/chapter/10-multiplication-first-steps#def-g2-mult-def) therefore forces at least one [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-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](#def-g8-equations-def) solved so far in this chapter was of first degree and had exactly one solution; a square in the [equation](#def-g8-equations-def) 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](#def-g8-equations-def) 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](#exo-g8-equations-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](https://one-course.com/books/math/1/en/chapter/3-subtraction-first-steps#ex-g1-subtraction-difference) of two consecutive squares is the [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) of the two numbers — the same $2n + 1$ as in question 8, now obtained from the third identity instead of the first.