0.1 Use the Language of Algebra

Aligned outcomes:

SLO 6.NS

The Number System. Divide fractions by fractions; compute fluently with

SLO 6.EE

Expressions and Equations. Write, read and evaluate expressions in which

SLO 7.NS

The Number System. Add, subtract, multiply and divide rational numbers,

SLO 7.EE

Expressions and Equations. Apply properties of operations to add, subtract,

SLO 8.NS

The Number System. Know that numbers which are not rational are irrational;

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:
  • read, write, and round whole numbers using place value;
  • distinguish a variable from a constant, and an expression from an equation;
  • translate an English phrase into an algebraic expression and read one back into English;
  • write repeated multiplication using exponent notation;
  • simplify an expression using the order of operations and evaluate it for a given value;
  • identify like terms and combine them.

People sometimes describe algebra as arithmetic with letters, which makes it sound like a small change. It is bigger than that. Arithmetic asks you to compute one answer from specific numbers. Algebra asks you to describe a relationship that holds no matter which numbers you use, and to do that you need a language precise enough that two people reading the same expression understand exactly the same thing.

This section builds that language. Most of it will look familiar from earlier grades, and that is fine — the goal here is not new material, it is precision. A surprising amount of trouble later in this course traces back to something small right here: an exponent attached to the wrong quantity, a minus sign attached to the wrong term, a phrase translated in the wrong order.

0.1.1 Whole Numbers, Place Value, and Rounding

The whole numbers are \(0, 1, 2, 3, 4, \dots\) — the counting numbers together with zero. They go on forever, so there is no largest one.

Our number system is positional, which means a digit's value depends on where it sits. The number 4,072 is not "four, zero, seven, two." It is

$$4{,}072 = 4(1000) + 0(100) + 7(10) + 2(1)$$

Reading right to left, the places are ones, tens, hundreds, thousands, ten-thousands, and so on, each one worth ten times the place to its right. Commas group digits in threes, which is why we say "four thousand, seventy-two."

That zero in the hundreds place is doing real work. Take it away and 4,072 collapses into 472. A zero that holds a place open is called a placeholder, and it is the reason the whole positional system works.

NumberDigit in the hundreds placeWhat that digit is worth
4,07200
4,5725500
41,07200
900,3183300
Place value in the number 4,072 A four-column chart. The digits 4, 0, 7 and 2 sit in the thousands, hundreds, tens and ones places. Each column shows what that digit is worth: 4 thousands is 4,000; the 0 in the hundreds place is worth nothing but holds the place open; 7 tens is 70; 2 ones is 2. The four values add to 4,072. thousands hundreds tens ones 4 0 7 2 4,000 0 70 2 + + + holds the place open 4,000 + 0 + 70 + 2 = 4,072

Figure 0.1.1 — Each digit of 4,072 sits in a place that fixes what it is worth.

To write a number in words, read it in comma-separated groups and name each group as you go: 68,204,915 is "sixty-eight million, two hundred four thousand, nine hundred fifteen."

Two conventions cause most of the mistakes here. English does not use "and" inside a whole number, so it is "two hundred four," not "two hundred and four" — we save the word "and" for the decimal point, which matters in Section 0.4. And a two-digit number between twenty-one and ninety-nine takes a hyphen whenever it is built from two words: "sixty-eight," "ninety-five." The multiples of ten in that range are single words and take no hyphen: "thirty," "ninety."

Rounding

Rounding swaps a number for a nearby one that is easier to work with, and you pay for that convenience with some accuracy. Here is the procedure:

Rounding a whole number.

1. Find the digit in the place you are rounding to.

2. Look at the digit immediately to its right.

3. If that digit is 5 or more, add 1 to the rounding digit. Otherwise leave it alone.

4. Replace every digit to the right with zeros.

Round 6,473 to the nearest hundred. The hundreds digit is 4, and the digit to its right is 7. Since 7 is 5 or more, the 4 becomes 5 and everything after it becomes zero, giving 6,500.

Round that same 6,473 to the nearest thousand instead. Now the rounding digit is 6, and the digit to its right is 4. Since 4 is less than 5, the 6 stays put and the rest go to zero, giving 6,000.

Always round from the original

Round 6,473 to the nearest hundred and you get 6,500. Round that result to the nearest thousand and you get 7,000 — but rounding 6,473 directly gives 6,000. Each rounding step adds a little error, and chaining them lets the error compound into a wrong answer.

One case surprises people: when the rounding digit is 9 and it has to increase, it rolls over into the next place. Rounding 3,962 to the nearest hundred turns the 9 into 10, which carries, giving 4,000.

Example 0.1.1: Rounding to two different places

Round 27,548 to the nearest thousand, then to the nearest ten thousand.

Solution

Step 1 — Nearest thousand. The thousands digit is 7. The digit to its right is 5, which is 5 or more, so the 7 becomes 8 and everything to the right becomes zero.

$$27{,}548 \rightarrow 28{,}000$$

Step 2 — Nearest ten thousand. Go back to the original number, 27,548. The ten-thousands digit is 2. The digit to its right is 7, which is 5 or more, so the 2 becomes 3.

$$27{,}548 \rightarrow 30{,}000$$

Notice we started from 27,548 both times rather than rounding 28,000 in the second step.

Answer: 28,000 and 30,000.

Try It Now 0.1.1

Round 8,461 to the nearest hundred, and round 5,097 to the nearest hundred.

Solution

8,461 to the nearest hundred. The hundreds digit is 4. The digit to its right is 6, which is 5 or more, so the 4 becomes 5.

$$8{,}461 \rightarrow 8{,}500$$

5,097 to the nearest hundred. The hundreds digit is 0. The digit to its right is 9, which is 5 or more, so the 0 becomes 1.

$$5{,}097 \rightarrow 5{,}100$$

Answer: 8,500 and 5,100.

0.1.2 Variables, Expressions, and Equations

Think about a phone plan that charges a flat $25 every month plus $10 for each gigabyte of data you use.

The $25 never changes. The number of gigabytes changes from month to month, and so does your bill. That difference has names.

Definition 0.1.1: Variable

A variable is a letter that represents a number whose value may change.

A variable is a letter whose value may change, so one expression covers every case The expression 25 plus 10 g stands at the top with the letter g in the accent color, above a note that g is the number of gigabytes used. Below a rule, three columns appear one after another and all remain on screen. The first sets g to 0, substitutes it as 25 plus 10 times 0, and gives a bill of 25 dollars. The second sets g to 3, substitutes 25 plus 10 times 3, and gives 55 dollars. The third sets g to 6, substitutes 25 plus 10 times 6, and gives 85 dollars. A closing note reads: one line, every month. The 25 and the 10 were the same in all three. a plan that charges $25 a month, plus $10 for each gigabyte of data used 25 + 10g may change g = 0 25 + 10(0) $25 g = 3 25 + 10(3) $55 g = 6 25 + 10(6) $85 One line, every month. The 25 and the 10 were the same in all three.

Definition 0.1.1 — Variable: one line covers every month, because the letter is the only thing that changes.

Definition 0.1.2: Constant

A constant is a number whose value always stays the same.

If we let \(g\) stand for gigabytes used, the monthly bill is

$$25 + 10g$$

What makes that line powerful is that it covers every possible month at once. It is not the bill for one particular month; it is the rule that produces the bill for any month, as soon as you know \(g\).

You can use any letter as a variable, though \(x\), \(y\), \(z\), \(a\), \(b\), and \(n\) are the usual choices. Picking a letter that hints at its meaning — \(g\) for gigabytes, \(t\) for time, \(C\) for cost — makes your own work much easier to read a week later.

A constant is a number whose value always stays the same Two expressions are annotated. In the first, 25 plus 10 g, a bar is drawn beneath the 25 and labelled the flat fee, the same every month, and a second bar is drawn beneath the 10 and labelled the rate per gigabyte, also unchanging; the g is left in the accent color, given its own accent bar, and labelled the only free part. In the second expression, 7 x squared plus 3 x minus 4, a bar is drawn beneath the minus 4 and labelled a constant term with no variable part at all, so it is its own coefficient. A closing line reads: a constant is a number whose value always stays the same. which numbers in an expression are settled before you start? 25 + 10 g 25: the flat fee, the same every month 10: the rate per gigabyte, also unchanging g: the only free part 7x2 + 3x − 4 −4 is a constant term: no variable part, so it is its own coefficient. A constant is a number whose value always stays the same.

Definition 0.1.2 — Constant: the numbers that are settled before you start, in both senses the section uses.

Operation symbols

Algebra adjusts a few arithmetic symbols, and both changes exist to prevent confusion.

Summary of operation notation. The table below collects the forms we will use from here on.

OperationArithmetic writesAlgebra prefersWhy the change
Addition\(a + b\)\(a + b\)no change needed
Subtraction\(a - b\)\(a - b\)no change needed
Multiplication\(3 \times 4\)\(3 \cdot 4\), \(3(4)\), or \(3b\)\(\times\) looks too much like the variable \(x\)
Division\(12 \div 3\)\(\dfrac{12}{3}\)the fraction bar also groups, which \(\div\) does not

Two habits follow from that table, and you will use both constantly. First, writing two things next to each other means multiply: \(5n\) is "5 times \(n\)," with no symbol between them because none is needed. This only works when a variable is involved — 54 is fifty-four, not five times four. Second, the fraction bar groups. In \(\dfrac{a+b}{2}\) the bar tells you to add first and then divide, whereas writing it with \(\div\) would force you to add parentheses: \((a+b) \div 2\).

Expressions and equations are different things

Definition 0.1.3: Expression

An expression is a combination of numbers, variables, and operation symbols that names a quantity. It contains no equals sign.

A phrase versus a sentence

An expression is a phrase — "the total cost" — so it names something but does not claim anything. An equation is a full sentence — "the total cost is 85 dollars" — so it makes a claim that can be true or false. You simplify a phrase; you solve a sentence.

An expression names a quantity, and contains no equals sign The expression 25 plus 10 g is broken into its parts by brackets drawn beneath it. The 25 and the 10 are bracketed and labelled numbers, the g is bracketed and labelled variable, and the plus sign is bracketed and labelled operation symbol. To the right, an empty dashed box stands where an equals sign would be, labelled no equals sign, nothing is claimed. Two more expressions from the section's translation table, 2 y minus 4 and 2 times open paren a plus 7 close paren, are colour keyed the same way beneath. A closing line reads: solve 25 plus 10 g does not mean anything, because there is no claim to make true. an expression names a quantity, and that is all it does 25 + 10 g number number variable operation symbol no equals sign nothing is claimed 2y − 4 2(a + 7) same three ingredients, and no equals sign in either “solve 25 + 10g” does not mean anything, because there is no claim to make true.

Definition 0.1.3 — Expression: numbers, variables and operation symbols naming a quantity, with no equals sign anywhere.

Definition 0.1.4: Equation

An equation is two expressions joined by an equals sign, asserting that they have the same value.

Comparing the two. Keeping these straight saves a lot of confusion later:

ExpressionEquation
A phrase. It names a quantity, like \(25 + 10g\).A sentence. It makes a claim, like \(25 + 10g = 85\).
No equals sign. Nothing is being asserted.Has an equals sign. Something is being asserted.
You simplify it. Rewrite it in a cleaner but equivalent form.You solve it. Find the values that make the claim true.
Result is a quantity. Another expression, or a number.Result is a value for the variable.

Asking someone to "solve" \(25 + 10g\) does not mean anything, because there is no claim to make true. Asking them to "simplify" \(25 + 10g = 85\) is confused in the same way.

You can see the difference in the phone plan. The expression \(25 + 10g\) answers "what is my bill?" The equation \(25 + 10g = 85\) asks something else entirely: "how much data would give me an $85 bill?"

An equation is a claim, and only one value of g makes it true A balance beam pivots on a central post. The left pan carries the expression 25 plus 10 g and the right pan carries 85. Trying g equals 3 puts 55 on the left, and the beam tips down to the right, labelled less than 85, false. Trying g equals 8 puts 105 on the left, and the beam tips down to the left, labelled more than 85, false. Trying g equals 6 puts 85 on the left and the beam comes level, labelled true, and it stays level for the rest of the loop. A closing line reads: you simplify an expression, you solve an equation, and solving means finding the value that makes the claim true. an equation is a claim, and a claim can be false 25 + 10g 85 g = 3 → 25 + 10(3) = 55 55 is less than 85 – false g = 8 → 25 + 10(8) = 105 105 is more than 85 – false g = 6 → 25 + 10(6) = 85 85 equals 85 – true You simplify an expression. You solve an equation – you find the value that makes the claim true.

Definition 0.1.4 — Equation: two expressions joined by an equals sign, making a claim that one value makes true.

Symbols for comparing

Not every relationship is an equality.

SymbolRead as
\(a = b\)\(a\) equals \(b\)
\(a \neq b\)\(a\) is not equal to \(b\)
\(a < b\)\(a\) is less than \(b\)
\(a > b\)\(a\) is greater than \(b\)
\(a \le b\)\(a\) is less than or equal to \(b\)
\(a \ge b\)\(a\) is greater than or equal to \(b\)

The strict symbols \(<\) and \(>\) leave equality out; \(\le\) and \(\ge\) include it. That distinction does real work in Chapter 9.

Parentheses \((\ )\), brackets \([\ ]\), and braces \(\{\ \}\) all group, and they all mean the same thing. The different shapes exist only so nested groupings stay readable, as in \(2[3 + 4(5 - 1)]\). The fraction bar and the radical sign group as well.

Try It Now 0.1.2

A gym charges a $40 sign-up fee plus $15 per month. Write an expression for the total cost after \(m\) months. Then write an equation stating that the total cost is $145. Which one would you use to find how many months you have been a member?

Solution

Step 1 — The expression. The $40 is paid once, so it is a constant. The $15 repeats every month, so it is multiplied by the number of months.

$$40 + 15m$$

Step 2 — The equation. Set that expression equal to the stated total.

$$40 + 15m = 145$$

Step 3 — Which one answers the question. The question gives you a total and asks for the number of months, so you need the claim that can be true or false — the equation. You would solve it for \(m\).

Answer: Expression \(40 + 15m\); equation \(40 + 15m = 145\); use the equation.

0.1.3 Translating Words into Algebra

A lot of what makes word problems hard is translation, not mathematics. Certain English phrases map onto operations reliably, and learning the map removes most of the difficulty.

OperationPhrases that signal it
Additionthe sum of; increased by; more than; total of; plus
Subtractionthe difference of; decreased by; less than; minus; subtracted from
Multiplicationthe product of; times; twice; of; multiplied by
Divisionthe quotient of; divided by; the ratio of; per
Equalsis; gives; yields; results in; will be

Word order is the trap

Addition and multiplication can be written in either order without changing the answer. Subtraction and division cannot — and English word order for those two is often reversed from the algebra.

"Less than" reads backwards

"5 less than \(n\)" is \(n - 5\), not \(5 - n\). The phrase names the amount being removed first and the starting quantity second, so the algebra comes out in the opposite order from the reading. "Subtracted from" flips it the same way.

Here is how those four phrases actually translate:

When you are unsure, substitute a number and check whether the result makes sense. If \(n = 12\), then "5 less than \(n\)" ought to be 7. Testing \(n - 5\) gives 7, which is right. Testing \(5 - n\) gives \(-7\), which is not.

EnglishAlgebra
the sum of \(x\) and 9\(x + 9\)
4 less than twice \(y\)\(2y - 4\)
the product of 6 and \(m\), decreased by 1\(6m - 1\)
the quotient of \(p\) and 3\(\dfrac{p}{3}\)
8 more than the product of 5 and \(k\)\(5k + 8\)
twice the sum of \(a\) and 7\(2(a + 7)\)

That last row shows why parentheses matter so much. "Twice the sum of \(a\) and 7" doubles the whole sum, so it is \(2(a+7)\). Without the parentheses, \(2a + 7\) would double only the \(a\) and leave the 7 alone — a different quantity. Any time an operation applies to a whole phrase rather than a single quantity, that phrase needs grouping symbols around it.

Reading algebra back into English

The translation runs both directions, and going backwards is a good test of whether you really understand an expression. Read the structure from the outside in: identify the last operation performed, name it first, then describe its pieces.

Take \(3(m - 2)\). The outermost operation is the multiplication by 3, and what it multiplies is a difference, so this is "three times the difference of \(m\) and 2." Compare it with \(3m - 2\), where the outermost operation is the subtraction: "two less than three times \(m\)." The two expressions differ by nothing but a pair of parentheses, and the English has to place the words in a different order to keep them apart.

Example 0.1.2: Turning a situation into an equation

A theater sells adult tickets for $12 and student tickets for $8. One night it takes in $960. Write an equation describing that night.

Solution

Step 1 — Name the changing quantities. Two things vary: how many of each ticket sold. Let \(a\) be the number of adult tickets and \(s\) the number of student tickets.

Step 2 — Build each piece of revenue. Each adult ticket brings in $12, so adult revenue is \(12a\). Each student ticket brings in $8, so student revenue is \(8s\).

Step 3 — Combine them. Total revenue is the sum:

$$12a + 8s$$

Step 4 — Turn the stated fact into a claim. The words "takes in $960" give us an amount to set the total equal to.

$$12a + 8s = 960$$

Answer: \(12a + 8s = 960\).

That example is the whole modeling process in miniature: quantities become variables, relationships become expressions, and a stated fact becomes an equation. Chapters 2 and 6 do exactly this at a larger scale.

Try It Now 0.1.3

Write an algebraic expression for each phrase: (a) 7 less than \(w\); (b) three times the sum of \(k\) and 2; (c) the quotient of 30 and \(n\), increased by 4.

Solution

(a) 7 less than \(w\). "Less than" reverses the reading order, so the 7 is removed from \(w\).

$$w - 7$$

(b) Three times the sum of \(k\) and 2. The tripling applies to the entire sum, so the sum needs parentheses.

$$3(k + 2)$$

(c) The quotient of 30 and \(n\), increased by 4. Take the quotient first, then add 4 to it.

$$\dfrac{30}{n} + 4$$

Answer: \(w - 7\); \(3(k+2)\); \(\dfrac{30}{n} + 4\).

0.1.4 Exponent Notation and the Order of Operations

Repeated multiplication gets its own shorthand. Rather than writing \(7 \cdot 7 \cdot 7 \cdot 7\), we write \(7^4\).

Definition 0.1.5: Exponential notation

For a counting number \(n\) — that is, \(n \ge 1\) — the expression \(a^n\) means the product of \(n\) factors of \(a\):

$$a^n = \underbrace{a \cdot a \cdot a \cdots a}_{n \text{ factors}}$$

The number \(a\) is the base, the repeated factor. The number \(n\) is the exponent, which counts how many times the base appears as a factor.

An exponent grabs only what it touches

In \(3x^2\) the exponent applies to \(x\) alone, so you square \(x\) and then multiply by 3. In \((3x)^2\) it applies to the whole product, giving \(9x^2\). The parentheses are the difference between two different quantities.

We read \(7^4\) as "seven to the fourth power." Two powers get special names because of where they come from geometrically: \(a^2\) is "\(a\) squared," from the area of a square, and \(a^3\) is "\(a\) cubed," from the volume of a cube.

The same care applies to minus signs, and this pair is worth memorizing:

ExpressionValueWhat the exponent applies to
\(-4^2\)\(-16\)just the 4; the negation happens afterward
\((-4)^2\)\(16\)the whole quantity \(-4\)
\(3x^2\)\(3 \cdot x \cdot x\)just the \(x\)
\((3x)^2\)\(9x^2\)the whole product \(3x\)

Chapter 1 comes back to exponents properly in Section 1.3.

The exponent counts the factors, and it reaches only what it touches Four sevens multiplied together sit under a bracket labelled four factors. An arrow leads down to the same quantity written as 7 with a raised 4, with the 7 labelled base, the repeated factor, and the raised 4 in accent labelled exponent, counting how many times the base is a factor, read as seven to the fourth power. Below a rule, two expressions are contrasted: negative 4 squared equals negative 16, with a short bracket under just the 4 showing the exponent reaches only the 4, and open paren negative 4 close paren squared equals 16, with a wide bracket under the whole negative 4 showing the parentheses hand the exponent the whole quantity. A footer repeats the same contrast in letters: 3 x squared is 3 times x times x, while open paren 3 x close paren squared is 9 x squared. repeated multiplication gets a shorthand, and the shorthand has a reach 7 · 7 · 7 · 7 4 factors 7 4 base – the repeated factor exponent – counts the factors read “seven to the fourth power” − 4 2 = −16 the exponent reaches only the 4 (−4) 2 = 16 the parentheses hand it the whole quantity the same in letters: 3x2 is 3 · x · x, while (3x)2 is 9x2

Definition 0.1.5 — Exponential notation: the exponent counts the factors, and it grabs only what it touches.

The order of operations

The expression \(2 + 3 \cdot 4\) could mean 20 or 14 depending on which operation you do first. Mathematics settles that with one fixed convention so every expression has exactly one value.

The order of operations.

1. Grouping symbols — parentheses, brackets, braces, fraction bars, radicals. Work from the innermost outward.

2. Exponents.

3. Multiplication and division — left to right, as they appear.

4. Addition and subtraction — left to right, as they appear.

Steps 3 and 4 are where most people misremember the rule. Multiplication does not outrank division, and addition does not outrank subtraction. Inside each of those two levels you simply work left to right:

$$20 \div 5 \cdot 2 = 4 \cdot 2 = 8$$ $$10 - 4 + 3 = 6 + 3 = 9$$

If you had done the multiplication before the division in the first line you would have gotten 2, and if you had added before subtracting in the second you would have gotten 3. Both are wrong.

Example 0.1.3: Working from the inside out

Simplify \(\;5 + 2\left[3^2 - (8 - 6)\right]\).

Solution

Step 1 — Innermost grouping. The parentheses are inside the brackets, so they go first.

$$5 + 2[3^2 - 2]$$

Step 2 — Exponent inside the brackets. Still working inside the grouping symbols.

$$5 + 2[9 - 2]$$

Step 3 — Finish the bracket.

$$5 + 2[7]$$

Step 4 — Multiplication before addition.

$$5 + 14$$

Step 5 — Addition.

$$19$$

Answer: 19.

Simplifying an expression in order-of-operations steps The expression 5 plus 2 times bracket 3 squared minus open paren 8 minus 6 close paren close bracket is simplified one step at a time. First the innermost grouping, 8 minus 6, becomes 2. Then the exponent 3 squared becomes 9. Then the bracket 9 minus 2 becomes 7. Then 2 times 7 becomes 14 by multiplying before adding. Finally 5 plus 14 is 19. Each line names the rule that produced it. 5 + 2[32 − (8 − 6)] the expression 5 + 2[32 − 2] innermost grouping first 5 + 2[9 − 2] then the exponent 5 + 2[7] finish the bracket 5 + 14 multiply before adding 19 add last

Figure 0.1.2 — One expression simplified a step at a time, each line naming the rule that produced it.

Evaluating an expression

To evaluate an expression, substitute a number for each variable and simplify what is left. Always wrap the substituted value in parentheses — this one habit prevents most sign errors.

Evaluate \(n^2 - 5n\) when \(n = -3\):

$$(-3)^2 - 5(-3) = 9 - (-15) = 9 + 15 = 24$$

Without those parentheses, \(-3^2\) would have been read as \(-9\), and the answer would have come out 6 instead of 24. Signed-number arithmetic is the subject of Section 0.2, and this is where careful notation starts paying for itself.

Terms and like terms

A term is a single number, a single variable, or a product of numbers and variables — the pieces an expression breaks into at its \(+\) and \(-\) signs. In \(7x^2 + 3x - 4\) the terms are \(7x^2\), \(3x\), and \(-4\). The number multiplying a variable is its coefficient, so here the coefficient of \(x^2\) is 7 and the coefficient of \(x\) is 3.

Two cases come up constantly and are worth naming now. When a variable appears with no number written in front of it, the coefficient is 1, because multiplying by 1 changes nothing: \(y\) means \(1y\), and \(-y\) means \(-1y\), so the coefficient of \(-y\) is \(-1\). And a term that is just a number, like the \(-4\) above, is called a constant term; it has no variable part, and the term is its own coefficient.

Definition 0.1.6: Like terms

Like terms are terms with the same variables raised to the same powers. Only like terms can be combined, and combining them means adding their coefficients.

$$7x^2 + 3x - 4 + 2x^2 + 5x = 9x^2 + 8x - 4$$

The \(7x^2\) and \(2x^2\) combine because both count "\(x\) squared" quantities, and \(3x\) and \(5x\) combine because both count "\(x\)" quantities.

Why \(9x^2\) and \(8x\) will not combine

They measure different kinds of thing. If \(x\) is a length, then \(x^2\) is an area — and there is no way to add an area to a length and report one number. No conversion exists, which is exactly why the two terms have to stay apart.

Constant terms are like terms with each other too, since none of them carries a variable. So in \(6b + 4 - 2b + 9\) the \(6b\) and \(-2b\) combine to \(4b\), and the constants 4 and 9 combine to 13, leaving \(4b + 13\). Watch the sign travelling with each term: the term after \(6b\) is \(-2b\), not \(2b\).

This is the simplification you will do more than any other for the rest of the course. Section 0.5 explains why it is allowed in the first place: combining like terms is the distributive property read backwards.

Like terms have the same variable raised to the same power, so their coefficients add The expression 7 x squared plus 3 x minus 4 plus 2 x squared plus 5 x is shown along the top. Tinted bands group its terms by kind: the two x squared terms in accent bands, the two x terms in curve bands, and the lone minus 4, its sign included, in a grey band. Below, the x squared terms combine as 7 x squared plus 2 x squared equals 9 x squared, the x terms combine as 3 x plus 5 x equals 8 x, and a note records that minus 4 has no like partner and travels down unchanged. Beneath a rule the combined result reads 9 x squared plus 8 x minus 4. A closing line reads: if x is a length, then x squared is an area, and there is no way to add an area to a length and report one number. only terms with the same variable part can be combined 7x2 + 3x − 4 + 2x2 + 5x 7x2 + 2x2 = 9x2 3x + 5x = 8x −4 is a constant term with no like partner, so it travels down unchanged 9x2 + 8x − 4 combining like terms means adding their coefficients if x is a length, then x2 is an area – and there is no way to add an area to a length and report one number

Definition 0.1.6 — Like terms: same variable, same power, so their coefficients add; everything else stays apart.

Try It Now 0.1.4

Simplify \(\;4 + 3(2^3 - 5)\), then evaluate \(\;x^2 - 2x\) when \(x = -4\).

Solution

Part 1 — Simplify \(4 + 3(2^3 - 5)\).

Start inside the parentheses. The exponent comes before the subtraction:

$$4 + 3(8 - 5)$$

Finish the grouping:

$$4 + 3(3)$$

Multiply before adding:

$$4 + 9 = 13$$

Part 2 — Evaluate \(x^2 - 2x\) at \(x = -4\).

Substitute with parentheses around the \(-4\):

$$(-4)^2 - 2(-4)$$

Square first, then multiply:

$$16 - (-8)$$

Subtracting a negative adds:

$$16 + 8 = 24$$

Answer: 13 and 24.

Problem Set 0.1

Problem 1. Write 52,806 in words.

Solution

Step 1 — Break the number into comma groups: 52,806 splits into two groups, 52 and 806, and each group gets named in turn.

Step 2 — Name each group in words: The first group, 52, is "fifty-two" — hyphenated because it is built from two words. The second group, 806, is "eight hundred six," with no "and" between the hundreds and the ones, since English saves "and" for the decimal point.

Answer: fifty-two thousand, eight hundred six.

Problem 2. Write "three hundred nine thousand, forty" in standard form.

Solution

Step 1 — Convert each named group back to digits: "Three hundred nine" fills the thousands group, giving 309. "Forty" fills the ones group, and since that group needs three digits, it is written 040.

Step 2 — Rejoin the groups with a comma: Placing the ones group directly after the thousands group gives the standard form.

$$309{,}040$$

Answer: 309,040.

Problem 3. Round 4,738 to the nearest hundred.

Solution

Step 1 — Find the hundreds digit and look one place right: In 4,738 the hundreds digit is 7, and the digit immediately to its right, in the tens place, is 3.

Step 2 — Decide whether to round up: Since 3 is less than 5, the hundreds digit stays 7 and everything to its right becomes zero.

$$4{,}738 \rightarrow 4{,}700$$

Answer: 4,700.

Problem 4. Round 4,738 to the nearest thousand.

Solution

Step 1 — Go back to the original 4,738, not the last answer: This time the rounding digit is the thousands digit, 4, and the digit to its right, in the hundreds place, is 7.

Step 2 — Decide whether to round up: Since 7 is 5 or more, the thousands digit increases from 4 to 5, and everything after it becomes zero.

$$4{,}738 \rightarrow 5{,}000$$

Answer: 5,000.

Problem 5. Round 19,952 to the nearest hundred.

Solution

Step 1 — Identify the hundreds digit and its neighbor: In 19,952 the hundreds digit is 9, and the tens digit to its right is 5.

Step 2 — Round up and follow the carry: Since 5 is 5 or more, the 9 has to increase by 1, but 9 + 1 is 10, so it rolls over: the hundreds place becomes 0 and the 1 carries into the thousands place. The thousands digit is also 9, so the same thing happens again — it becomes 0 and carries into the ten-thousands place, turning that 1 into a 2.

Step 3 — Zero out the rest: Every digit to the right of the hundreds place becomes zero.

$$19{,}952 \rightarrow 20{,}000$$

Answer: 20,000.

Problem 6. In the number 736,215, what is the value of the digit 3?

Solution

Step 1 — Locate the digit 3 and name its place: Reading 736,215 from the right, the places are ones (5), tens (1), hundreds (2), thousands (6), ten-thousands (3), and hundred-thousands (7). The digit 3 sits in the ten-thousands place.

Step 2 — Multiply the digit by its place value: A digit in the ten-thousands place is worth that many ten-thousands.

$$3 \times 10{,}000 = 30{,}000$$

Answer: 30,000.

Problem 7. A bakery charges a $20 setup fee plus $4 per dozen cookies. Write an expression for the cost of \(d\) dozen cookies.

Solution

Step 1 — Separate the fixed charge from the charge that repeats: The $20 setup fee is paid once no matter how many cookies are ordered, so it is a constant. The $4 charge applies to every dozen, so it must be multiplied by the number of dozens, \(d\).

Step 2 — Combine the two pieces: Add the constant setup fee to the repeating cost.

$$20 + 4d$$

Answer: \(20 + 4d\).

Problem 8. Using your expression from the previous problem, write an equation stating that the total cost is $68.

Solution

Step 1 — Take the expression as the left side: The expression \(20 + 4d\) from the previous problem already names the total cost for \(d\) dozen cookies.

Step 2 — Assert the stated total with an equals sign: Turning the expression into a claim requires setting it equal to the given amount, $68.

$$20 + 4d = 68$$

Answer: \(20 + 4d = 68\).

Problem 9. Explain in one sentence why \(6x - 1\) is an expression and \(6x - 1 = 11\) is an equation.

Solution

Answer: \(6x - 1\) has no equals sign, so it only names a quantity — a phrase you can simplify but not solve — while \(6x - 1 = 11\) has an equals sign that asserts a claim about that quantity, a sentence you can solve for the value of \(x\) that makes it true.

Problem 10. Translate into algebra: the sum of \(t\) and 15.

Solution

Step 1 — Identify the operation word: "Sum" always signals addition, and addition can be written in either order without changing the value.

$$t + 15$$

Answer: \(t + 15\).

Problem 11. Translate into algebra: 9 less than \(m\).

Solution

Step 1 — Watch the order: "Less than" reads backwards — the amount named first is the amount being taken away, and the quantity it is taken away from comes second in the sentence but first in the algebra. So "9 less than \(m\)" removes 9 from \(m\), not the other way around.

$$m - 9$$

Step 2 — Check the reversal with a number: If \(m = 20\), "9 less than \(m\)" should be 11. Testing \(m - 9\) gives \(20 - 9 = 11\), which is right; testing \(9 - m\) would give \(-11\), which is not.

Answer: \(m - 9\).

Problem 12. Translate into algebra: twice the sum of \(r\) and 5.

Solution

Step 1 — Build the sum first: "The sum of \(r\) and 5" names a single quantity, \(r + 5\).

Step 2 — Apply the doubling to that whole quantity: "Twice" multiplies by 2, and since it multiplies the entire sum rather than just \(r\), the sum needs parentheses around it.

$$2(r + 5)$$

Answer: \(2(r + 5)\).

Problem 13. Translate into algebra: the quotient of 24 and \(y\), decreased by 3.

Solution

Step 1 — Take the quotient first: "The quotient of 24 and \(y\)" divides 24 by \(y\), giving \(\dfrac{24}{y}\).

Step 2 — Decrease that quotient by 3: "Decreased by 3" subtracts 3 from whatever came before it, so the 3 attaches to the outside of the quotient, not inside it.

$$\dfrac{24}{y} - 3$$

Answer: \(\dfrac{24}{y} - 3\).

Problem 14. Translate into English: \(4(n + 6)\).

Solution

Step 1 — Name the outermost operation: Reading from the outside in, \(4(n+6)\) is a product: 4 times something, and that something is a sum.

Step 2 — Describe the piece being multiplied: The quantity in parentheses is "the sum of \(n\) and 6," and the 4 multiplies that entire sum, which is why the parentheses are there.

Answer: Four times the sum of \(n\) and 6.

Problem 15. Write \(5 \cdot 5 \cdot 5 \cdot 5 \cdot 5\) using exponent notation.

Solution

Step 1 — Count the repeated factor: The base being multiplied is 5, and it appears as a factor 5 times.

$$5 \cdot 5 \cdot 5 \cdot 5 \cdot 5 = 5^5$$

Answer: \(5^5\).

Problem 16. Evaluate \(2^5\).

Solution

Step 1 — Write out the repeated factors: \(2^5\) means 2 multiplied by itself 5 times.

$$2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$$

Step 2 — Multiply left to right:

$$2 \cdot 2 = 4, \quad 4 \cdot 2 = 8, \quad 8 \cdot 2 = 16, \quad 16 \cdot 2 = 32$$

Answer: 32.

Problem 17. Explain why \(-6^2\) and \((-6)^2\) have different values, and give both values.

Solution

Step 1 — Identify what the exponent touches in \(-6^2\): There are no parentheses grouping the negative sign with the 6, so the exponent applies only to 6; the negation happens after the squaring is done.

$$-6^2 = -(6^2) = -(36) = -36$$

Step 2 — Identify what the exponent touches in \((-6)^2\): Here the parentheses group \(-6\) as a single quantity, so the exponent applies to the whole thing, negative sign included.

$$(-6)^2 = (-6)(-6) = 36$$

Step 3 — State why they differ: The two expressions square different quantities — one squares just 6 and negates afterward, the other squares \(-6\) itself — and squaring a negative number always gives a positive result, while negating a positive square keeps it negative.

Answer: \(-6^2 = -36\) and \((-6)^2 = 36\); they differ because the parentheses decide whether the negative sign is part of what gets squared.

Problem 18. Simplify \(\;18 \div 3 \cdot 2\).

Solution

Step 1 — Recognize the precedence tier: Division and multiplication sit on the same level of the order of operations, so neither outranks the other — work left to right through the expression as written.

Step 2 — Divide first, since it comes first reading left to right:

$$18 \div 3 = 6$$

Step 3 — Multiply next:

$$6 \cdot 2 = 12$$

Answer: 12.

Problem 19. Simplify \(\;15 - 6 + 2\).

Solution

Step 1 — Work left to right: Addition and subtraction sit at the same level in the order of operations, so there is no "do addition first" rule to reach for — you simply take the operations in the order they're written.

$$15 - 6 + 2 = 9 + 2$$

Step 2 — Finish the addition:

$$9 + 2 = 11$$

Answer: \(11\).

Problem 20. Simplify \(\;7 + 2(4^2 - 9)\).

Solution

Step 1 — Exponent first, inside the parentheses: Grouping symbols are handled from the inside out, and inside this one an exponent still outranks the subtraction next to it.

$$7 + 2(4^2 - 9) = 7 + 2(16 - 9)$$

Step 2 — Finish the parentheses:

$$7 + 2(7)$$

Step 3 — Multiply before adding:

$$7 + 14$$

Step 4 — Add:

$$21$$

Answer: \(21\).

Problem 21. Simplify \(\;3[10 - 2(1 + 2)]\).

Solution

Step 1 — Innermost grouping first: The parentheses sit inside the brackets, so the addition inside them goes before anything else.

$$3[10 - 2(1 + 2)] = 3[10 - 2(3)]$$

Step 2 — Multiply inside the bracket: Multiplication still outranks subtraction even while it's trapped inside a bracket.

$$3[10 - 6]$$

Step 3 — Finish the bracket:

$$3[4]$$

Step 4 — Multiply:

$$12$$

Answer: \(12\).

Problem 22. Evaluate \(\;a^2 + 3a\) when \(a = 5\).

Solution

Step 1 — Substitute \(a = 5\), parentheses around the substituted value: Wrapping the value in parentheses keeps the exponent and the sign attached to it, exactly where they belong.

$$a^2 + 3a = (5)^2 + 3(5)$$

Step 2 — Apply the exponent, then multiply:

$$25 + 15$$

Step 3 — Add:

$$40$$

Answer: \(40\).

Problem 23. Evaluate \(\;a^2 + 3a\) when \(a = -2\).

Solution

Step 1 — Substitute \(a = -2\), keeping it in parentheses: This is the case the parentheses habit is built for — without them, the exponent could grab the wrong thing.

$$a^2 + 3a = (-2)^2 + 3(-2)$$

Step 2 — Square first, since \((-2)^2\) squares the whole quantity in parentheses, then multiply:

$$4 + (-6)$$

Step 3 — Add the signed values:

$$4 - 6 = -2$$

Answer: \(-2\).

Problem 24. Identify the terms and the coefficient of each term in \(\;9y^2 - y + 12\).

Solution

Step 1 — Split the expression at its \(+\) and \(-\) signs: A term is whatever sits between those signs, so \(9y^2 - y + 12\) breaks into three pieces: \(9y^2\), \(-y\), and \(12\).

Step 2 — Read off the coefficient of each term: A coefficient is the number multiplying the variable part. The term \(9y^2\) has coefficient \(9\). The term \(-y\) means \(-1y\), since no number is written in front of a lone variable, so its coefficient is \(-1\). The constant term \(12\) has no variable part at all, so by convention it is its own coefficient.

Answer: Terms \(9y^2\), \(-y\), and \(12\), with coefficients \(9\), \(-1\), and \(12\).

Problem 25. Combine like terms: \(\;6b + 4 - 2b + 9\).

Solution

Step 1 — Group the like terms: Terms with the same variable to the same power can combine, so the \(b\)-terms gather together and the constant terms gather together.

$$6b + 4 - 2b + 9 = (6b - 2b) + (4 + 9)$$

Notice the sign attached to \(2b\) travels with it — the term right after \(6b\) is \(-2b\), not \(2b\).

Step 2 — Add the coefficients within each group:

$$4b + 13$$

Answer: \(4b + 13\).

Problem 26. Combine like terms: \(\;5m^2 + 3m - m^2 + 7m\).

Solution

Step 1 — Sort the terms by variable part: \(5m^2\) and \(-m^2\) both count "\(m\) squared" quantities, while \(3m\) and \(7m\) both count plain "\(m\)" quantities, so those are the two groups that are allowed to combine.

$$5m^2 + 3m - m^2 + 7m = (5m^2 - m^2) + (3m + 7m)$$

Step 2 — Combine each group: Remember \(-m^2\) means \(-1m^2\), so \(5m^2 - m^2\) is \(5m^2 - 1m^2\).

$$4m^2 + 10m$$

Answer: \(4m^2 + 10m\).

Problem 27. Explain why \(4x^2\) and \(4x\) are not like terms.

Solution

Answer: \(4x^2\) and \(4x\) are not like terms because their variable parts are raised to different powers — \(x^2\) and \(x\) — and like terms require the same variable to the same power, not just the same variable. If \(x\) is a length, \(x^2\) measures an area, and there is no way to add an area to a length and report a single number, so the two terms have to stay separate however far the simplifying goes.

Key Terms

whole numbers — the counting numbers together with zero: 0, 1, 2, 3, and so on.

placeholder — a zero that holds a place open so the remaining digits keep their values.

rounding — replacing a number with a nearby, simpler number, at the cost of some accuracy.

variable — a letter that represents a number whose value may change.

constant — a number whose value always stays the same.

expression — a combination of numbers, variables, and operations that names a quantity; it has no equals sign.

equation — two expressions joined by an equals sign, asserting they have the same value.

base — in \(a^n\), the factor \(a\) that is repeated.

exponent — in \(a^n\), the number \(n\) counting how many times the base is used as a factor.

term — a number, a variable, or a product of them; the pieces an expression separates into at its \(+\) and \(-\) signs.

coefficient — the number multiplying the variable part of a term.

like terms — terms with the same variables raised to the same powers, and therefore combinable.