Primary & Middle School Mathematics · Grades 1–9

66Algebra and Equations

Letters stand for numbers, and computing with letters proves facts about all numbers at once. This chapter practices expanding and factoring — including the three identities that will follow you through all of mathematics — and solves first-degree equations, product equations, and simple inequalities, always one justified step at a time.

66.1 Expanding

Definition 66.1 (Expanding)

Expanding means transforming a product into a sum, using distributivity:

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

Example 66.2

Every term of the first bracket multiplies every term of the second:

\begin{align*} (x + 3)(2x + 5) &= x \times 2x + x \times 5 + 3 \times 2x + 3 \times 5 \\ &= 2x^2 + 5x + 6x + 15 \\ &= 2x^2 + 11x + 15 . \end{align*}

Signs travel with the terms:

\begin{align*} (3x - 2)(x - 4) &= 3x^2 - 12x - 2x + 8 = 3x^2 - 14x + 8 . \end{align*}

Theorem 66.3 (The three identities)

For all numbers $a$ and $b$ :

(a+b)^2 = a^2 + 2ab + b^2, \qquad (a-b)^2 = a^2 - 2ab + b^2, \qquad (a+b)(a-b) = a^2 - b^2 .

Proof. Expand each left-hand side. For the first: $(a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$ . For the second, replace $b$ by $-b$ in the first. For the third: $(a+b)(a-b) = a^2 - ab + ba - b^2 = a^2 - b^2$ (the cross terms cancel). ∎

The identity (a+b)2 = a2 + 2ab + b2, seen as areas: the big square of side a+b is made of two squares and two equal rectangles. Forgetting the “2ab” is forgetting the two rectangles! — The identity $(a+b)^2 = a^2 + 2ab + b^2$ , seen as areas: the big square of side $a+b$ is made of two squares and two equal rectangles. Forgetting the “ $2ab$ ” is forgetting the two rectangles!

Example 66.4

(x + 5)^2 = x^2 + 10x + 25, \qquad (3x - 1)^2 = 9x^2 - 6x + 1, \qquad (2x + 7)(2x - 7) = 4x^2 - 49 .

Mental arithmetic: $102^2 = (100 + 2)^2 = 10000 + 400 + 4 = 10404$ , and $98 \times 102 = (100-2)(100+2) = 10000 - 4 = 9996$ .

66.2 Factoring

Method 66.5 (Factoring)

To factor an expression, try in order:

common factor: $ka + kb = k(a + b)$ , where $k$ may itself be a bracket;
difference of squares: $a^2 - b^2 = (a+b)(a-b)$ ;
perfect square: $a^2 + 2ab + b^2 = (a+b)^2$ (and with $-2ab$ , $(a-b)^2$ ).

Always check by expanding the result.

Example 66.6

Common factor: $15x^2 + 10x = 5x \times 3x + 5x \times 2 = 5x(3x + 2)$ .

Common bracket:

(x+2)(x-1) + (x+2)(3x+4) = (x+2)\bigl[(x-1) + (3x+4)\bigr] = (x+2)(4x+3).

Difference of squares: $x^2 - 81 = (x+9)(x-9)$ , and

(2x+1)^2 - 9 = (2x+1)^2 - 3^2 = (2x+1+3)(2x+1-3) = (2x+4)(2x-2).

Each factor still has a common factor: $(2x+4)(2x-2) = 2(x+2) \times 2(x-1) = 4(x+2)(x-1)$ .

66.3 Equations

Method 66.7 (First-degree equations)

To solve an equation like $5x - 3 = 2x + 9$ :

gather the $x$ -terms on one side, the numbers on the other, by adding the same quantity to both sides;
reduce each side;
divide both sides by the coefficient of $x$ ;
check the solution in the original equation.

Example 66.8

\begin{align*} 5x - 3 &= 2x + 9 \\ 5x - 2x &= 9 + 3 && \text{(add $3 - 2x$ to both sides)}\\ 3x &= 12 \\ x &= 4 && \text{(divide by 3).} \end{align*}

Check: $5 \times 4 - 3 = 17$ and $2 \times 4 + 9 = 17$ . The solution is $4$ .

With brackets, expand first: $3(x - 2) = x + 8$ becomes $3x - 6 = x + 8$ , so $2x = 14$ and $x = 7$ .

Theorem 66.9 (Zero-product rule)

A product is zero if and only if at least one factor is zero:

A \times B = 0 \quad\Longleftrightarrow\quad A = 0 \ \text{ or } \ B = 0 .

Proof. If a factor is zero the product clearly is. Conversely if $A B = 0$ with $A \neq 0$ , dividing both sides by $A$ gives $B = 0$ . ∎

Example 66.10

Solve $(x - 3)(2x + 10) = 0$ : either $x - 3 = 0$ or $2x + 10 = 0$ , so the solutions are $3$ and $-5$ .

Solve $x^2 - 49 = 0$ : factor first, $(x-7)(x+7) = 0$ , solutions $7$ and $-7$ .

Solve $x^2 = 3x$ : never divide by $x$ ! Move and factor: $x^2 - 3x = x(x - 3) = 0$ , solutions $0$ and $3$ .

66.4 Inequalities

Proposition 66.11 (Rules for inequalities)

An inequality is preserved when adding the same number to both sides, and when multiplying both sides by the same positive number. Multiplying both sides by a negative number reverses the inequality sign.

Proof. If $a \leq b$ then $b - a \geq 0$ . Adding $c$ : $(b+c) - (a+c) = b - a \geq 0$ , so $a + c \leq b + c$ . Multiplying by $c > 0$ : $bc - ac = (b-a)c \geq 0$ , so $ac \leq bc$ . Multiplying by $c < 0$ : $(b-a)c \leq 0$ , so $ac \geq bc$ : the order is reversed. ∎

Example 66.12

Solve $7 - 2x < 15$ :

\begin{align*} 7 - 2x &< 15 \\ -2x &< 8 && \text{(subtract 7)} \\ x &> -4 && \text{(divide by $-2$: reverse the sign).} \end{align*}

The solutions are all numbers greater than $-4$ : on a number line, a hollow dot at $-4$ and shading to the right.

The solutions of 7 - 2x < 15: all x > -4 (hollow dot: -4 itself is not a solution). — The solutions of $7 - 2x < 15$ : all $x > -4$ (hollow dot: $-4$ itself is not a solution).

66.5 Exercises

Exercise 66.1 ★

Expand and reduce:

4(3x - 2), \qquad (x + 6)(x + 2), \qquad (2x - 3)(x + 5), \qquad (x - 1)(x - 9).

Solution

Solution of Exercise 66.1.

$4(3x - 2) = 12x - 8$ .

$(x+6)(x+2) = x^2 + 2x + 6x + 12 = x^2 + 8x + 12$ .

$(2x-3)(x+5) = 2x^2 + 10x - 3x - 15 = 2x^2 + 7x - 15$ .

$(x-1)(x-9) = x^2 - 9x - x + 9 = x^2 - 10x + 9$ .

Exercise 66.2 ★

Expand using the identities:

(x + 8)^2, \qquad (3x - 4)^2, \qquad (x + 10)(x - 10), \qquad (5x + 2)(5x - 2).

Solution

Solution of Exercise 66.2.

$(x+8)^2 = x^2 + 16x + 64$ .

$(3x-4)^2 = 9x^2 - 24x + 16$ .

$(x+10)(x-10) = x^2 - 100$ .

$(5x+2)(5x-2) = 25x^2 - 4$ .

Exercise 66.3 ★

Compute mentally, using an identity: $101^2$ ; $59^2$ (write $59 = 60 - 1$ ); $53 \times 47$ .

Solution

Solution of Exercise 66.3.

$101^2 = (100+1)^2 = 10000 + 200 + 1 = 10201$ .

$59^2 = (60-1)^2 = 3600 - 120 + 1 = 3481$ .

$53 \times 47 = (50+3)(50-3) = 2500 - 9 = 2491$ .

Exercise 66.4 ★

Factor:

9x + 12, \qquad x^2 - 64, \qquad x^2 + 14x + 49, \qquad 18x^2 - 6x .

Solution

Solution of Exercise 66.4.

$9x + 12 = 3(3x + 4)$ .

$x^2 - 64 = (x+8)(x-8)$ .

$x^2 + 14x + 49 = (x+7)^2$ .

$18x^2 - 6x = 6x(3x - 1)$ .

Exercise 66.5 ★

Solve, writing every step:

4x + 7 = 19, \qquad 6x - 5 = 2x + 11, \qquad 5(x - 3) = 3x + 1 .

Solution

Solution of Exercise 66.5.

$4x + 7 = 19$ : $4x = 12$ , so $x = 3$ .

$6x - 5 = 2x + 11$ : $4x = 16$ , so $x = 4$ .

$5(x-3) = 3x + 1$ : $5x - 15 = 3x + 1$ , so $2x = 16$ and $x = 8$ .

Exercise 66.6 ★★

Solve using the zero-product rule:

(x + 4)(3x - 12) = 0, \qquad x^2 - 121 = 0, \qquad (2x - 1)^2 = 0, \qquad x^2 = 8x .

Solution

Solution of Exercise 66.6.

$(x+4)(3x-12) = 0$ : $x = -4$ or $x = 4$ .

$x^2 - 121 = (x-11)(x+11) = 0$ : $x = 11$ or $x = -11$ .

$(2x-1)^2 = 0$ : the single solution $x = \frac12$ .

$x^2 = 8x$ : $x(x - 8) = 0$ , so $x = 0$ or $x = 8$ .

Exercise 66.7 ★★

Factor $E = (3x + 2)^2 - 25$ , then solve $E = 0$ .

Solution

Solution of Exercise 66.7.

Difference of squares with $a = 3x + 2$ and $b = 5$ :

E = (3x + 2)^2 - 5^2 = (3x + 2 + 5)(3x + 2 - 5) = (3x + 7)(3x - 3).

Zero-product rule: $3x + 7 = 0$ gives $x = -\frac73$ , and $3x - 3 = 0$ gives $x = 1$ .

Exercise 66.8 ★★

Solve the inequalities and represent the solutions on a number line:

3x - 4 \geq 8, \qquad 5 - 2x > 1, \qquad 4(x + 1) \leq x - 5 .

Solution

Solution of Exercise 66.8.

$3x - 4 \geq 8$ : $3x \geq 12$ , so $x \geq 4$ (filled dot at $4$ , shade right).

$5 - 2x > 1$ : $-2x > -4$ , dividing by $-2$ reverses the sign: $x < 2$ (hollow dot at $2$ , shade left).

$4(x+1) \leq x - 5$ : $4x + 4 \leq x - 5$ , so $3x \leq -9$ , so $x \leq -3$ (filled dot at $-3$ , shade left).

Exercise 66.9 ★★

I think of a number, multiply it by $3$ , add $7$ , and get the same result as if I had multiplied it by $5$ and subtracted $9$ . What is my number? (Set up an equation and solve it.)

Solution

Solution of Exercise 66.9.

Let $x$ be the number: $3x + 7 = 5x - 9$ . Then $16 = 2x$ , so $x = 8$ . Check: $3 \times 8 + 7 = 31$ and $5 \times 8 - 9 = 31$ .

Exercise 66.10 ★★★

Let $n$ be a whole number.

Expand $(n + 1)(n - 1)$ and deduce that the product of the two neighbors of $n$ is always $n^2 - 1$ (e.g. $24 \times 26 = 25^2 - 1 = 624$ ).
Show that the sum of three consecutive integers is always a multiple of $3$ .

Solution

Solution of Exercise 66.10.

1. $(n+1)(n-1) = n^2 - 1$ (third identity). The neighbors of $n$ are $n - 1$ and $n + 1$ , so their product is $n^2 - 1$ : e.g. $24 \times 26 = 625 - 1 = 624$ .

2. Three consecutive integers can be written $n - 1$ , $n$ , $n + 1$ . Their sum is

(n-1) + n + (n+1) = 3n ,

a multiple of $3$ (three times the middle one).

66.6 Problem: The algebra of digits

Problem 66.1

Weekend problem — why the divisibility rules work, the 1089 trick unmasked, and mental arithmetic that looks like sorcery

Grade school taught you rules (a number is divisible by $9$ when its digit sum is); magicians perform tricks (everyone ends on $1089$ ); markets know shortcuts ( $35^2 = 1225$ , instantly). All of it is the same secret: a number with digits $a$ , $b$ , $c$ is the expression $100a + 10b + c$ , and this chapter’s algebra (Theorem 66.3, expanding, factoring) can take it apart. By the end of this problem you will have proved the rules, unmasked the trick, and learned the sorcery.

Part I — Digits under the microscope.

A two-digit number with tens digit $a$ and units digit $b$ is $10a + b$ . Expand and factor the difference between the number and its reversal, $(10a + b) - (10b + a)$ , and state the discovery. Check it on $72 - 27$ .
Prove the rule of nine for three-digit numbers: verify the identity

$100a + 10b + c = 9\,(11a + b) + (a + b + c),$

and explain why it shows that a number and its digit sum leave the same remainder in the division by $9$ — so one is divisible by $9$ exactly when the other is (Proposition 64.3, now proved).
Deduce the rule of three from the same identity, in one sentence.
“Casting out nines”, the accountants’ check: to test $47 \times 86 = 4\,042$ , replace each number by the digit sum of its digit sum: $47 \to 11 \to 2$ , $86 \to 14 \to 5$ , and $2 \times 5 = 10 \to 1$ ; then $4\,042 \to 10 \to 1$ : consistent. Explain why a correct product must pass the test (write $47 = 9k + 2$ , $86 = 9l + 5$ and expand), and find a wrong result that the test fails to catch (why are swapped digits invisible to it?).
Prove the rule of eleven for four-digit numbers: verify

$1000a + 100b + 10c + d = 11\,(91a + 9b + c) + (- a + b - c + d),$

and use it to decide whether $2\,926$ is divisible by $11$ .

Part II — The 1089 trick. The trick: think of a three-digit number whose first and last digits differ by at least $2$ ; reverse it; subtract the smaller from the larger; reverse the result; add the last two numbers. The magician announces: 1089.

Perform the trick on $742$ and on one number of your choice.
Let the number have digits $a$ , $b$ , $c$ with $a > c$ . Factor the first subtraction, $(100a + 10b + c) - (100c + 10b + a)$ , and list all the values it can take when $a - c$ runs from $2$ to $9$ .
Each of those multiples of $99$ has a striking digit pattern. Writing $m = a - c$ , verify the identity

$99m = 100\,(m - 1) + 90 + (10 - m),$

and read off the three digits of $99m$ .
Add $99m$ to its own reversal, using question 8’s digits, and prove that the total is $1089$ — whatever $m$ is. The trick is dead; long live the algebra.
Why did the magician require the first and last digits to differ by at least $2$ ? Examine the cases $a = c$ and $a - c = 1$ (where the subtraction gives $99$ , which must be written $099$ for the reversal to work), and state the small print that keeps the trick honest.

Part III — Sorcery for daily use.

The birthday trick: “multiply your birth month by $5$ , add $7$ , double, add your birth day, subtract $14$ .” Show that the result is always $10m + d$ — month and day, readable at a glance. (Try it on someone.)
You redesign the trick as: “multiply the month by $5$ , add $9$ , double, add the day.” What number must the spectator now subtract at the end for the same month–day readout?
The missing-digit trick: the spectator picks any number, subtracts its digit sum (question 2 says the result is always a multiple of $9$ — explain why for a three-digit number), then crosses out one nonzero digit and reads you the remaining digits. Explain how you recover the crossed-out digit, and why $0$ had to be excluded.
Market sorcery: prove the identity $(10t + 5)^2 = 100\,t(t+1) + 25$ , state the rule for squaring any number ending in $5$ , and compute $45^2$ , $85^2$ and $105^2$ in your head.
Finale, extending Exercise 66.10: prove that the sum of five consecutive integers is always a multiple of $5$ , but that the sum of four consecutive integers is never a multiple of $4$ . (Two short computations — and one moral about what algebra does to the words “always” and “never”.)

Solution

Solution of Problem 66.1.

1. $(10a + b) - (10b + a) = 9a - 9b = 9(a - b)$ : the difference between a two-digit number and its reversal is always a multiple of $9$ . Check: $72 - 27 = 45 = 9 \times 5$ , and $a - b = 7 - 2 = 5$ .

2. Expanding the right side: $9(11a + b) + a + b + c = 99a + 9b + a + b + c = 100a + 10b + c$ . So the number equals a multiple of $9$ plus its digit sum: dividing by $9$ , both leave the same remainder. In particular remainder $0$ for one means remainder $0$ for the other: the rule of nine, proved.

3. $9(11a + b)$ is also a multiple of $3$ , so the same identity shows number and digit sum leave the same remainder in the division by $3$ : the rule of three.

4. Write $47 = 9k + 2$ and $86 = 9l + 5$ . Then

47 \times 86 = (9k + 2)(9l + 5) = 9\,(9kl + 5k + 2l) + 10 :

the product leaves the same remainder as $2 \times 5 = 10$ , namely $1$ . A correct result must therefore also leave remainder $1$ — which $4\,042$ does. But the test only sees remainders: $4\,042$ and $4\,402$ (digits swapped) pass identically, so passing proves nothing — failing is what convicts.

5. Expanding: $11(91a + 9b + c) - a + b - c + d = 1001a + 99b + 11c - a + b - c + d = 1000a + 100b + 10c + d$ . So the number is divisible by $11$ exactly when its alternating sum $-a + b - c + d$ is. For $2\,926$ : $-2 + 9 - 2 + 6 = 11$ , divisible by $11$ : yes, $2\,926 = 11 \times 266$ .

6. $742 - 247 = 495$ ; $495 + 594 = 1\,089$ . Any valid example lands on $1\,089$ too.

7. $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$ . For $a - c = 2, 3, \dots, 9$ :

198,\ 297,\ 396,\ 495,\ 594,\ 693,\ 792,\ 891 .

8. $100(m - 1) + 90 + (10 - m) = 100m - 100 + 90 + 10 - m = 99m$ . So the three digits of $99m$ are $m - 1$ , then $9$ , then $10 - m$ (check $495$ : $m = 5$ gives $4$ , $9$ , $5$ ).

9. The reversal of $99m$ has digits $10 - m$ , $9$ , $m - 1$ , so it equals $100(10 - m) + 90 + (m - 1)$ . Adding:

100\,(m - 1 + 10 - m) + 180 + (10 - m + m - 1) = 100 \times 9 + 180 + 9 = 1\,089 ,

independent of $m$ . Whatever the starting number, the magician is safe.

10. If $a = c$ , the subtraction gives $0$ and the trick dies at once. If $a - c = 1$ , it gives $99$ , which must be treated as the three-digit string $099$ (reversal $990$ , $99 + 990 = 1\,089$ : saved!). The small print: differences of $1$ work only if the spectator writes the subtraction result with three digits, leading zero included — requiring a gap of at least $2$ avoids the argument.

11. $2(5m + 7) + d - 14 = 10m + 14 + d - 14 = 10m + d$ : the tens (and hundreds, for October–December) show the month, the units (and tens) the day.

12. $2(5m + 9) + d = 10m + 18 + d$ : subtract $18$ .

13. By question 2, $n - (\text{digit sum of } n) = 9(11a + b)$ for a three-digit $n$ : always a multiple of $9$ (similar identities work for any length). A multiple of $9$ has digit sum a multiple of $9$ (question 2 again), so the digits the spectator reads out must be completed to the next multiple of $9$ : the missing digit is that complement. Ambiguity: if the read digits already sum to a multiple of $9$ , the crossed digit could be $0$ or $9$ — excluding $0$ removes the doubt.

14. $(10t + 5)^2 = 100t^2 + 100t + 25 = 100\,t(t + 1) + 25$ : write $t(t+1)$ , then glue $25$ behind it. $45^2$ : $4 \times 5 = 20$ , so $2\,025$ . $85^2$ : $8 \times 9 = 72$ , so $7\,225$ . $105^2$ : $10 \times 11 = 110$ , so $11\,025$ .

15. Five consecutive: $n + (n+1) + (n+2) + (n+3) + (n+4) = 5n + 10 = 5(n + 2)$ : a multiple of $5$ , always. Four consecutive: $n + (n+1) + (n+2) + (n+3) = 4n + 6 = 4(n + 1) + 2$ : remainder $2$ in the division by $4$ , never a multiple. Algebra converts “it seems to work” into always, and “I found no example” into never — two words no amount of testing can reach (Method 48.8).