Variables as placeholders, the four operations in algebraic shorthand, the write-substitute-simplify routine for evaluating an expression, substituting into the formulas for distance and perimeter, and writing an expression of your own from a described situation.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 1 — Connections to Algebra
Variables in Algebra
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-8 — the lesson these objectives are drawn from
Warm-up
You have been evaluating expressions since long before anyone used the word. Naming the routine is what makes it reliable.
Discussion prompt
A ticket costs 12 dollars. Work out the cost for 3 people, then for 7 people. What stayed the same between those two calculations, and what changed?
Hint: Write both calculations out fully. The parts that match are the interesting part.
Answer:
\[ 3 \text{ people:} \; 12 \cdot 3 = 36 \qquad 7 \text{ people:} \; 12 \cdot 7 = 84 \]
The rule stayed the same — twelve dollars times the number of people — and only the number of people changed. Algebra gives that unchanging rule a name of its own by writing the changing part as a letter: twelve times n. You have already been doing the hard part; the letter is just bookkeeping.
Concept
In algebra a letter stands for a number you either do not know yet or do not want to fix. Write the rule once with the letter in it, and it covers every case at the same time instead of one case at a time.
variable — A letter used to represent a range of numbers. The numbers it can represent are called its values.
That is why a single line of algebra can replace a whole table of arithmetic.
Figure (svg): A box labelled t that can hold any of the numbers 1, 2, 3 or 4, with the expression 180 times t beside it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-3
Section
Section 1
Concept
An expression is a legal combination of numbers, letters and operations. If a letter appears anywhere in it, it is a variable expression. If not, it is a numerical expression and it stands for one specific number.
variable expression — An expression built from constants, variables and operations, such as 180t or a plus b plus c.
Figure (svg): Two columns contrasting numerical expressions, which contain only numbers, with variable expressions, which contain at least one letter
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-3 — the Assigning Variables opening paragraphs
Picture it
The reason a letter is worth the trouble: it says four things at once.
Figure (svg): Four bars showing the value of 180t growing as t takes the values 1, 2, 3 and 4 hours
Every bar comes from the same expression, 180t. What changed from bar to bar is the value of t, never the rule itself. Hold on to that and the rest of the chapter is easy.
Worked example
Guided Practice 1 to 4. Trivial once you know what you are looking for, and worth being certain about.
\[ \text{Identify the variable in each: } \; y + 15, \quad 20 - s, \quad \tfrac{10}{b}, \quad rt. \]
Scan each expression for letters
Why: A variable is just a letter standing where a number could stand. Numbers and operation signs are never variables.
In y plus 15 the letter is y
Why: Fifteen is a fixed number, so it cannot vary. The letter is the only part that can.
In 20 minus s the letter is s
Why: Same reading. The twenty is fixed.
In ten over b the letter is b
Why: The fraction bar is a division sign; it is an operation, not a variable.
In rt there are two variables, r and t
Why: Two letters written side by side means they are multiplied, and both of them can vary.
Figure (svg): The solution to Worked example name the variable shown as a ladder of expressions, one row per algebraic move
\[ y, \quad s, \quad b, \quad r \text{ and } t \]
Verify: count the letters in each expression
Why: One letter in each of the first three, and two letters in the last one — which matches, because rt is a product of two separate quantities, a rate and a time.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 6-6
Sorting
One question decides every one of these: is there a letter anywhere in it?
Sort into buckets
Drop each expression into the right column.
Notice that the two columns are the same four expressions before and after substitution. Substituting is the move that carries you from left to right.
Worked example
This is Example 1 from the textbook. Read each one out loud before you write anything.
\[ \text{State the meaning and the operation: } \; 8y, \quad \tfrac{16}{b}, \quad 4 + s, \quad 9 - x. \]
8y means 8 times y, and the operation is multiplication
Why: A number written directly against a letter always means multiply. The same product may be written 8 times y or as 8 and y each in brackets.
\[ 8 y =\text{ multiplication} \]
Sixteen over b means 16 divided by b, and the operation is division
Why: The fraction bar is a division sign. It can equally be written with the division symbol.
\[ \frac{16}{b} =\text{ division} \]
4 plus s means 4 plus s, and the operation is addition
Why: Addition is the one operation algebra never hides; the plus sign is always written.
\[ 4 + s =\text{ addition} \]
9 minus x means 9 minus x, and the operation is subtraction
Why: Subtraction is also always written out. Order matters here in a way it does not for the first two.
\[ 9 - x =\text{ subtraction} \]
Figure (svg): The solution to Worked example say what the shorthand means shown as a ladder of expressions, one row per algebraic move
\[ 8y: \text{ multiply} \quad \tfrac{16}{b}: \text{ divide} \quad 4+s: \text{ add} \quad 9-x: \text{ subtract} \]
Verify: substitute a number and see whether the size behaves
Why: Put two in for every letter: 8y becomes 16, which is bigger, as multiplying should be. Sixteen over b becomes 8, which is smaller, as dividing should be. The sizes move the way the named operations predict, so the readings are right.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-3
Trap
\[ 8y \]
Read 8y as the two-digit number eighty-something, or as 8 next to y with no operation
Why: Written arithmetic has trained you that digits side by side make one number, so 8y looks like a single object.
With that reading, evaluating at y equal to 2 produces nonsense such as eighty-two, and every later step inherits the error.
\[ 8y = 8 \cdot y = (8)(y) \]
Read every gap between a number and a letter as a times sign
Why: Algebra drops the times sign on purpose, because a raised dot is easily confused with a decimal point and the letter x is already taken.
\[ \text{At } y = 2: \quad 8y = 8(2) = 16 \]
The missing sign is the single most common misreading in the whole chapter. When in doubt, write the dot back in on your own paper.
Elimination
All four of these are written in correct algebra. Only one of them is a numerical expression.
Eliminate the wrong options
Which of these is a numerical expression?
Survives elimination: B
Why: Both factors in (8)(15) are numbers, so the expression can be simplified right now to the single number 120. Brackets written against each other mean multiply, exactly as in (8)(y), but here there is no letter left to wait for.
Notation
One short expression, carrying four separate conventions. Take them apart one at a time.
Annotate
On: \( 180t \)
Four conventions in five characters. This density is exactly why algebra is worth learning, and exactly why it has to be read slowly at first.
Prediction
Commit to an answer before you compute anything.
Predict first
The expression 180t is evaluated at t equal to 2 and then at t equal to 3. What happens to the value?
Correct: It rises from 360 to 540.
\[ 180t \big|_{t=2} = 180(2) = 360 \qquad 180t \big|_{t=3} = 180(3) = 540 \]
Why: The rule stays fixed while the variable moves, and that is the whole point of writing it with a letter. Substituting 2 gives 180 times 2, which is 360; substituting 3 gives 180 times 3, which is 540. The jump is 180, one extra copy of the rate, because one extra hour was travelled.
Section
Section 2
Concept
Addition and subtraction always show their signs. Multiplication and division frequently do not: multiplication may be written as nothing at all, and division is usually written as a fraction bar. Almost every early misreading comes from those two.
Figure (svg): Four rows pairing an algebra shorthand with the operation it means: 8y is multiplication, 16 over b is division, 4 plus s is addition, 9 minus x is subtraction
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-3 — Example 1, Describe the Variable Expression
Picture it
Cover the right-hand column and read each symbol aloud. Then cover the left and write the symbol from the meaning.
Figure (svg): Four rows pairing an algebra shorthand with the operation it means: 8y is multiplication, 16 over b is division, 4 plus s is addition, 9 minus x is subtraction
The first row is the one that costs marks. A number written against a letter, or a bracket written against a bracket, is always a multiplication.
Worked example
The textbook makes this point in a single line beside Example 3. It is worth slowing down on.
\[ \text{Write the product of } r \text{ and } t \text{ in every standard form.} \]
Write the letters side by side
Why: This is the compact form and the one you will meet most often in formulas.
Write them with a raised dot between
Why: Used when running the letters together would be ambiguous, for example when one of them is a whole word such as rate.
\[ r \cdot t \]
Write each in its own set of brackets
Why: Used when a value is being substituted, because it keeps a negative number safely fenced in.
\[ (r) (t) \]
Check that no form uses the multiplication cross
Why: The cross is avoided in algebra because it is indistinguishable from the letter x at speed.
Figure (svg): The solution to Worked example write the same product three ways shown as a ladder of expressions, one row per algebraic move
\[ rt \;=\; r \cdot t \;=\; (r)(t) \]
Verify: substitute the same pair of numbers into all three
Why: Taking r as 180 and t as 2, all three forms give 180 times 2, which is 360. Three notations, one number, so they really are the same product written differently.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 4-4
Matching
Four expressions, four operation names. Some of them are deliberately close together.
Match the pairs
Why: Brackets written against brackets mean multiply, so the first is a product of 8 and x. A fraction bar is a division sign, so the second is p divided by 4. The last two differ only in one character, and that character is the whole difference between adding and subtracting — which is why the sign has to be copied carefully every single time.
Worked example
Two of these four expressions change when you swap the two quantities, and two do not. Decide before you read the steps.
\[ \text{Compare } \; 4 + s \text{ with } s + 4, \quad \text{and } 9 - x \text{ with } x - 9. \]
Substitute a test number, say s equal to 6 and x equal to 6
Why: One well-chosen number settles a question like this far faster than an argument does.
\[ \text{let } s = 6, x = 6 \]
Evaluate both addition forms
Why: Four plus six and six plus four both come to ten, so swapping made no difference.
\[ 4 + 6 = 10, 6 + 4 = 10 \]
Evaluate both subtraction forms
Why: Nine minus six is three, but six minus nine is negative three. Swapping changed the answer.
\[ 9 - 6 = 3, 6 - 9 = -3 \]
State the conclusion carefully
Why: Addition may be written in either order. Subtraction may not, so the wording of a problem fixes which number goes first.
Figure (svg): The solution to Worked example does order matter here shown as a ladder of expressions, one row per algebraic move
\[ 4 + s = s + 4 \qquad \text{but} \qquad 9 - x \neq x - 9 \]
Verify: try a second pair of numbers
Why: With s and x both equal to 2, addition gives 6 either way, while subtraction gives 7 one way and negative 7 the other. A second test agreeing with the first is decent evidence, and Chapter 2 will prove it.
Trap
\[ 9 - x \quad \text{at } x = 4 \]
Read it as the larger number minus the smaller, giving 5 either way
Why: Arithmetic practice makes taking the small from the large feel automatic, so the written order gets overruled.
The habit is invisible while every answer is positive. It shows up as soon as x is bigger than 9, and then every answer has the wrong sign.
\[ 9 - x \quad \text{at } x = 4 \quad \text{and at } x = 12 \]
Substitute into the written order, whatever the sizes are
Why: The expression says nine minus x, so the x is always the one being taken away.
\[ 9 - 4 = 5 \qquad 9 - 12 = -3 \]
The second answer is negative, and that is correct rather than a mistake. Chapter 2 is entirely about numbers on that side of zero.
Translation
Each phrase on the left is written in ordinary English. Pair it with the algebra that says the same thing.
Match the pairs
Why: These are Guided Practice items 6 to 9. Two of the four hide their operation sign: five times c is written by putting the number against the letter, and p divided by four is written as a fraction. The two that keep their signs, subtraction and addition, are the two where order matters or where the sign is the only clue.
Two truths and a lie
Three of these four expressions are the same product written in different notations.
Eliminate the wrong options
Which one is NOT equal to the others?
Survives elimination: D
Why: The first three are the three standard ways of writing a product: letters run together, a raised dot, or separate brackets. The fourth uses a division bar, which is a completely different operation, and substituting any value other than one will separate it from the rest immediately.
Fill the middle
Each line names the hidden operation for you. Supply the value it produces at the given number.
Fill in the blanks
8y = 8 \cdot y, \; \text16 y = 2, \; 8y = 8 \qquad \tfrac______ = 16 \div b, \; \text___ b = 2, \; \tfrac______ = ___
Why: A number written directly against a letter is a product, so 8y at y equal to 2 is eight times two, or 16. A fraction bar is a division sign written vertically, so sixteen over b at b equal to 2 is sixteen divided by two, or 8. The two expressions use the same digits and land on different answers, which is exactly what makes reading the hidden sign correctly worth the effort.
Section
Section 3
Concept
To evaluate a variable expression you write the expression, substitute a number for each variable, and simplify. The number you end up with is called the value of the expression.
evaluate — To substitute a number for each variable in an expression and then simplify the result. The number produced is the value of the expression.
Figure (svg): Three labelled boxes in a row - write the expression, substitute the number, simplify - with 5y becoming 5 times 2 becoming 10
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 4-4 — the Evaluating Expressions key-concept box
Picture it
Most lost marks in this lesson come from doing the first and third box in your head and never writing the middle one.
Figure (svg): Three labelled boxes in a row - write the expression, substitute the number, simplify - with 5y becoming 5 times 2 becoming 10
The middle box is the one a marker can check and the one you can check tomorrow. Write it down even when the arithmetic is easy.
Worked example
This is Example 2 from the textbook, all four parts. The routine is identical every time.
\[ \text{Evaluate at } y = 2: \quad 5y, \quad \tfrac{10}{y}, \quad y + 6, \quad 14 - y. \]
Substitute 2 into 5y and simplify
Why: Five against the letter is a product, so this becomes five times two.
\[ 5(2) = 10 \]
Substitute 2 into ten over y and simplify
Why: The fraction bar divides, so this becomes ten divided by two.
\[ \frac{10}{2} = 5 \]
Substitute 2 into y plus 6 and simplify
Why: Straight addition, with the letter first this time.
\[ 2 + 6 = 8 \]
Substitute 2 into 14 minus y and simplify
Why: The y is what is being taken away, so it goes in the second position.
\[ 14 - 2 = 12 \]
Figure (svg): The solution to Worked example evaluate four expressions at y equal to 2 shown as a ladder of expressions, one row per algebraic move
\[ 5y = 10, \quad \tfrac{10}{y} = 5, \quad y + 6 = 8, \quad 14 - y = 12 \]
Verify: check each answer against the operation's expected direction
Why: Multiplying by five made the value larger than two; dividing ten by two made it smaller than ten; adding six moved up from two; subtracting two moved down from fourteen. Each result sits on the side the operation predicts, so none of the four was substituted into the wrong slot.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 4-4
Faded example
The first two lines are done. Supply the last one.
Fill in the blanks
5y \;=\; 5(2) \;=\; 10
Why: The substitution line has already replaced y with 2, so nothing algebraic is left to do. Five times two is ten, and ten is therefore the value of the expression 5y when y equals 2. Writing the substitution as its own line is what makes this final step a piece of pure arithmetic.
Worked example
Guided Practice 5 to 8. Do all four on paper before advancing, and write the middle line every time.
\[ \text{Evaluate at } x = 3: \quad 7x, \quad 5 + x, \quad \tfrac{12}{x}, \quad x - 2. \]
Write each expression down first
Why: Copying the expression before substituting is what stops a 7x turning into a 2x halfway through.
\[ 7 x, 5 + x, \frac{12}{x}, x - 2 \]
Substitute 3 everywhere, in brackets
Why: Brackets are not needed for a positive number here, but the habit pays for itself the moment the number is negative.
\[ 7(3), 5 + 3, \frac{12}{3}, 3 - 2 \]
Simplify each numerical expression
Why: Now there are no letters left, so ordinary arithmetic finishes the job.
\[ 21, 8, 4, 1 \]
Attach nothing else
Why: The value of an expression is a number. There is no unit here because the problem gave none.
Figure (svg): The solution to Worked example the same routine at x equal to 3 shown as a ladder of expressions, one row per algebraic move
\[ 7x = 21, \quad 5 + x = 8, \quad \tfrac{12}{x} = 4, \quad x - 2 = 1 \]
Verify: substitute the answers back into a size check
Why: Twenty-one is seven copies of three, four is how many threes fit in twelve, and both of the one-step answers moved by exactly the amount added or subtracted. Every value behaves the way its operation predicts.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 6-6
Error analysis
The student was asked to evaluate ten over y and 14 minus y at y equal to 2. Both answers are wrong, and for two different reasons.
Annotate
On: \( \tfrac{10}{y} = \tfrac{10}{2} = 20 \qquad 14 - y = 2 - 14 = -12 \)
The fix for both is the same discipline — write the substitution line without doing any arithmetic on it, then simplify that line on its own.
Ranking
Four moves, one correct order. Three of them are the official routine and one is a check.
Put in order
Why: Write, substitute, simplify is the textbook's three-step routine, and the size check belongs at the end where it can catch a slip. Doing the check earlier is useless because there is no number yet to check, and doing the substitution before writing the expression is how a 7x quietly becomes a 2x.
Prediction
You can often tell the direction of an answer before you compute it.
Predict first
The expression twelve over x is evaluated at x equal to 3 and then at x equal to 6. What happens to the value?
Correct: It halves, from 4 to 2.
\[ \tfrac{12}{x} \big|_{x=3} = 4 \qquad \tfrac{12}{x} \big|_{x=6} = 2 \]
Why: The variable is in the denominator, so making it larger cuts the value down rather than building it up. Twelve divided by three is four; twelve divided by six is two. Doubling what you divide by always halves the result, and noticing that before computing is a genuine check on the arithmetic afterwards.
Socratic
The arithmetic in this lesson is easy enough to do in your head. That is exactly why this question is worth answering now.
Discussion prompt
Give two concrete reasons for writing the substitution line down, even when you can see the answer immediately. At least one reason should be about a mistake it prevents.
Hint: Think about what you would look at if your answer disagreed with the back of the book.
Answer:
First, it separates two different kinds of mistake. If the substitution line is right and the final number is wrong, the error is arithmetic. If the substitution line itself is wrong, the error is algebraic — a misread operation or a swapped order. Without that line, both look identical.
Second, the expressions get longer. By Lesson 1.3 you will be substituting into expressions with several operations, and by Chapter 3 into equations with the letter on both sides. The habit has to be built while it is easy, because it cannot be built while it is hard.
Section
Section 4
Concept
A formula states a relationship that always holds — distance is rate times time, perimeter is the sum of the side lengths. Using one is exactly the evaluation routine you just learned, with one extra step at the end: attach the unit.
\[ d = rt \qquad P = a + b + c \]
The unit is not decoration. An answer of forty with no unit is not an answer to a question about a triangle in feet.
Figure (svg): A function machine labelled 180 times t with 2 going in and 360 coming out
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 4-5 — Examples 3 and 4
Picture it
Values go in at the left, the rule does its work, the answer comes out at the right.
Figure (svg): A function machine labelled 180 times t with 2 going in and 360 coming out
This picture is worth keeping: in Lesson 1.8 the same machine reappears under the name function, and nothing about it changes except the vocabulary.
Worked example
Example 3 from the textbook. Notice that the formula is written with letters first and numbers second.
\[ \text{A race car averages } 180 \text{ miles per hour. Find the distance travelled in } 2 \text{ hours.} \]
Write the formula
Why: Starting from the general rule rather than the numbers makes the substitution visible and checkable.
\[ d = r t \]
Substitute 180 for r and 2 for t
Why: The rate is the miles-per-hour figure and the time is the number of hours; matching them to the right letters is the only real decision here.
\[ d = 180(2) \]
Simplify
Why: Now it is ordinary arithmetic.
\[ d = 360 \]
Attach the unit
Why: Miles per hour multiplied by hours leaves miles, so the answer is a distance.
\[ d = 360\text{ miles} \]
Figure (svg): The solution to Worked example how far does the race car go shown as a ladder of expressions, one row per algebraic move
\[ d = rt = 180(2) = 360 \text{ miles} \]
Verify: check the unit and the size
Why: Miles per hour times hours cancels the hours and leaves miles, which is the unit a distance should carry. And 360 miles in two hours is 180 miles in one, which is the rate we started from — so the answer folds back into the question correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 4-4
Hypothesis
Set it up with the formula before you touch the numbers.
Predict first
A cyclist rides at an average speed of 14 miles per hour for 3 hours. Using the formula d equals rt, how far does she travel?
Correct: 42 miles.
\[ d = rt = 14(3) = 42 \text{ miles} \]
Why: Substituting into d equals rt gives 14 times 3, which is 42, and miles per hour times hours leaves miles. Riding for three hours at fourteen miles in each hour must give something near forty, so the size is right as well as the arithmetic. The three wrong options each replace the multiplication with a different operation, and each one fails a units check before the arithmetic is even reached: a speed cannot be added to, or subtracted from, a number of hours.
Worked example
Example 4 from the textbook, the geometry link. Same routine, three variables instead of two.
\[ \text{A triangle has sides } a = 8, \; b = 15, \; c = 17 \text{ feet. Find its perimeter.} \]
Write the formula
Why: Perimeter is the distance all the way round, which for a triangle is the sum of the three sides.
\[ P = a + b + c \]
Substitute the three side lengths
Why: Because the operation is addition, it does not matter which side is called which — a useful thing to notice rather than to assume.
\[ P = 8 + 15 + 17 \]
Simplify
Why: Eight and fifteen make twenty-three, and seventeen more makes forty.
\[ P = 40 \]
Attach the unit
Why: Every side was measured in feet, so the total is in feet.
\[ P = 40\text{ feet} \]
Figure (svg): A triangle with sides labelled 8, 15 and 17 feet, and the sum 8 plus 15 plus 17 equals 40 written beside it
\[ P = a + b + c = 8 + 15 + 17 = 40 \text{ feet} \]
Verify: re-add the sides in a different order
Why: Seventeen plus fifteen is thirty-two, plus eight is forty — the same total reached by a different route, which is a real check rather than a repetition of the same keystrokes.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 5-5
Trap
\[ d = rt = 180(2) = 360 \]
Stop at 360 and move to the next question
Why: The arithmetic is finished, so the work feels finished.
Three hundred and sixty what? The number alone does not answer a question about a car, and on a test it usually does not score full marks.
\[ d = rt = 180(2) = 360 \text{ miles} \]
Read the units of the quantities you substituted, and carry them through
Why: Miles per hour multiplied by hours leaves miles. The unit is determined by the formula, not chosen by you.
The habit is worth more than the mark it protects: if the surviving unit is not the unit the question asked for, the setup was wrong and you have just caught it before the marker did.
Missing information
A question can be perfectly well written and still be unanswerable.
Discussion prompt
A train travels at 60 miles per hour. How far does it go? Say exactly what one piece of information is missing, and what unit that missing quantity must be measured in for the formula to work.
Hint: Write the formula out and see which letter has no number against it.
Answer:
\[ d = rt, \quad r = 60, \quad t = \; ? \]
The time is missing. In the formula d equals rt there are three letters and only one of them has been given a value, so no distance can be produced. The missing quantity has to be a time measured in hours, because the rate is given per hour — if it arrived in minutes it would have to be converted first, or the answer would come out sixty times too large.
Comparison
Fill the blanks from memory. The pattern in the last column is the thing worth carrying forward.
Comparison matrix
| Formula | What the letters mean | Unit of the answer |
|---|---|---|
| d = rt | rate times time | miles |
| P = a + b + c | the three side lengths added | feet |
The unit of the answer is decided by the formula and by the units that went in — never by preference. Multiplying miles per hour by hours leaves miles; adding three lengths in feet leaves feet.
Estimation
A rough answer first is the cheapest error check there is.
Predict first
A car travels at 58 miles per hour for 4 hours. Roughly how far does it go?
Correct: A bit over 200 miles.
\[ d = rt = 58(4) = 232 \text{ miles} \]
Why: Rounding 58 up to 60 gives 60 times 4, which is 240, so the true answer is a little under 240 — in fact 232. Estimating first means that a slipped decimal point or a mistyped digit stands out immediately, because the exact answer has to land near the estimate.
Section
Section 5
Concept
So far the expressions have been handed to you. Writing your own is two decisions: choose a letter for the quantity that changes, say in words what is being done to it, and then write that in symbols.
A letter chosen to mean something, such as t for time or n for number of people, is easier to check a week later than a letter chosen at random.
Figure (svg): A box labelled t that can hold any of the numbers 1, 2, 3 or 4, with the expression 180 times t beside it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-3
Picture it
The reason to bother writing your own expression rather than just working out the case in front of you.
Figure (svg): Four bars showing the value of 180t growing as t takes the values 1, 2, 3 and 4 hours
Four rows of arithmetic collapse into one line of algebra. That compression is what you are buying, and it is why the letter is worth the extra thought.
Worked example
The race-car situation, but this time you write the algebra rather than being given it.
\[ \text{A car travels at } 180 \text{ miles per hour. Write an expression for the distance after } t \text{ hours.} \]
Decide what varies
Why: The speed is fixed at 180 for this problem. The number of hours is the thing that can change, so that is the variable.
\[ t =\text{ number of hours} \]
Say the rule in words
Why: Distance is the speed multiplied by the number of hours travelled.
\[ 180 \times t \]
Write it in symbols
Why: A number against a letter is a product, so no times sign is needed.
\[ 180 t \]
Test it on a case you can check by hand
Why: In one hour the car covers 180 miles, and the expression gives 180 times one, which is 180. The expression agrees with the obvious case.
\[ \text{at } t = 1\text{ gives } 180 \]
Figure (svg): The solution to Worked example write the expression from a description shown as a ladder of expressions, one row per algebraic move
\[ 180t \quad \text{miles after } t \text{ hours} \]
Verify: test a second, different value
Why: At t equal to 3 the expression gives 540, and three hours at 180 miles each really is 540 miles. Two agreeing cases is good evidence that the expression, and not just one answer, is right.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-4
Translation
Four situations, four expressions. Two of them contain a fixed part as well as a varying one.
Match the pairs
Why: The first two differ only in the fixed fee, and the fee is added rather than multiplied because it is paid once. Sharing means dividing, with the number of people underneath because that is what the total is being split among. Five fewer than n is n minus 5, not 5 minus n — the order in the English is the reverse of the order in the symbols, which is exactly why that phrase is worth practising.
Worked example
This one has two pieces, which is the shape most real situations take.
\[ \text{A club charges a } 20 \text{ dollar joining fee plus } 5 \text{ dollars per month. Write an expression for the cost after } m \text{ months.} \]
Separate the fixed part from the varying part
Why: Twenty dollars is paid once and never again, so it does not involve the variable at all. Five dollars is paid per month, so it does.
\[ \text{fixed } 20,\text{ varying } 5\text{ per month} \]
Write the varying part first
Why: Five dollars in each of m months is five multiplied by m, written without a times sign.
\[ 5 m \]
Add the fixed part
Why: The joining fee is added on once, not once per month, so it sits outside the product.
\[ 5 m + 20 \]
Check the wording against the symbols
Why: Reading it back: five times the number of months, plus twenty. That is exactly what the club charges.
\[ 5 m + 20 \]
Figure (svg): The solution to Worked example a cost with a fixed part and a varying part shown as a ladder of expressions, one row per algebraic move
\[ \text{cost} = 5m + 20 \text{ dollars after } m \text{ months} \]
Verify: test the case you can count on your fingers
Why: After one month the cost should be the twenty pound fee plus one month at five, which is 25. The expression gives 5 times 1 plus 20, which is 25. After zero months it gives 20, the joining fee alone, which is also right.
Trap
\[ \text{cost} = (5 + 20)m = 25m \]
Add the two dollar amounts first, then multiply the total by the number of months
Why: Both numbers are in dollars, so they look like they belong together.
\[ \text{At } m = 1: \; 25 \qquad \text{At } m = 2: \; 50 \]
The second month has quietly charged the joining fee a second time. The first case looked right, which is what makes this error so easy to keep.
\[ \text{cost} = 5m + 20 \]
Ask of each number: is this paid once, or once per month?
Why: Only the amounts paid per month may be multiplied by the number of months.
\[ \text{At } m = 1: \; 25 \qquad \text{At } m = 2: \; 30 \]
Testing at two different values is what separates the right expression from the wrong one. One test case is never enough when a fixed amount is involved.
Elimination
A taxi charges 3 dollars to get in, then 2 dollars for every mile travelled.
Eliminate the wrong options
Which expression gives the fare for a journey of m miles?
Survives elimination: A
Why: The two dollars is charged per mile, so it multiplies m. The three dollars is charged once, so it is added on outside. Testing at m equal to zero settles it instantly: a journey of no miles should cost the 3 dollar boarding charge, and only the first expression gives that.
Socratic
Any letter would work mathematically. Not every letter works equally well for a human being.
Discussion prompt
You are writing an expression for the total mass of n identical boxes, each of mass 4 kilograms. Why might a teacher prefer n or b to a randomly chosen letter here — and which single letter would you actively avoid in handwritten algebra?
Hint: Think about what the letter has to survive: your own handwriting, a week later.
Answer:
A letter chosen as an initial — n for number, b for boxes, t for time, m for months — is self-documenting. When you come back to 4n a week later you can still read what it counts, which matters more as expressions grow longer.
The letter to avoid where possible is x when a multiplication cross is anywhere nearby, because handwritten they are almost identical. That collision is the historical reason algebra dropped the cross in favour of writing symbols against each other in the first place.
Edge cases
Testing an expression at zero is fast and catches a surprising number of mistakes.
Discussion prompt
For the club cost 5m plus 20, what does the expression give when m is zero, and what does that number mean in the story? Then say what the wrong version 25m gives at zero, and why that answer is impossible.
Hint: What has happened in the real situation at the moment zero months have passed?
Answer:
\[ 5(0) + 20 = 20 \qquad 25(0) = 0 \]
At zero months you have joined and paid the joining fee but no monthly payments, so the cost should be exactly 20 dollars — which is what the correct expression gives. The wrong version gives 0, claiming the club is free until the first month elapses, and that contradicts the story directly.
Substituting zero isolates the fixed part of any expression, which makes it the single most useful test value when you are checking work of this kind.
Comparison
Fill the blanks from memory before you scroll back. The pattern in the middle column is the one that costs marks when it is missed.
Comparison matrix
| Operation | How algebra writes it | Does order matter? |
|---|---|---|
| Addition | 4 + s | No |
| Subtraction | 9 - x | Yes |
| Multiplication | 8y, 8 · y, or (8)(y) | No |
| Division | 16 over b, or 16 ÷ b | Yes |
The two operations that hide their signs are also, by coincidence, one order-free and one order-sensitive. So there is no shortcut: the sign has to be read, and then the order has to be read.
Pattern
Whether the question says evaluate, use the formula, or write an expression, the same five moves cover it.
Steps three and four are the pair most students merge into one, and merging them is why an arithmetic slip and an algebra slip become impossible to tell apart.
OpenStax Elementary Algebra 2e, §1.2 Use the Language of Algebra §1.2
Check
Reading the shorthand. Solve it before you click.
Check your understanding
What operation does the expression 7k represent?
Answer: B
Why: A number written directly against a letter always means multiply — algebra leaves the times sign out on purpose, because a cross is easily confused with the letter x. So 7k is seven times k, and at k equal to 3 it evaluates to 21.
Check
Evaluating. Write the substitution line before you choose.
Check your understanding
Evaluate the expression 11 minus k when k equals 3.
Answer: C
Why: Substituting gives eleven minus three, which is eight. The k is what is being taken away, so it stays in second position after the substitution — writing the line 11 minus 3 before simplifying is what keeps that order intact.
Check
Using a formula. Set it up in symbols first.
Check your understanding
A rectangle has a length of 9 metres and a width of 4 metres. Using the formula that perimeter equals twice the length plus twice the width, what is the perimeter?
Answer: A
Why: Substituting gives twice nine plus twice four, which is eighteen plus eight, or twenty-six metres. Every side was measured in metres and perimeter is a distance around the shape, so the answer carries metres rather than square metres.
Real world
A phone plan costs 15 dollars a month plus 8 cents for every minute of calls. You want a single expression that gives the monthly bill in dollars.
Discussion prompt
Choose your variable and say what it counts, then write the expression. Say which of the two numbers must be converted before the expression can be written, and why leaving it unconverted would be wrong rather than merely untidy.
Hint: Look hard at the units of the two amounts before you combine them.
Answer:
\[ \text{let } m = \text{ minutes of calls in the month} \]
\[ \text{bill} = 0.08m + 15 \text{ dollars} \]
The eight cents has to become 0.08 dollars first. Adding a quantity in cents to a quantity in dollars produces a number that is neither — it would claim that a hundred minutes of calls costs 815 dollars, or alternatively that the monthly fee is fifteen cents.
This is the same discipline as attaching a unit to a formula answer, applied one step earlier: quantities can only be added when they are measured in the same unit.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is the expression 5 minus n the same as n minus 5, and is 5n the same as n times 5?
Correct: 5n equals n times 5, but 5 minus n does not equal n minus 5.
\[ \text{At } n = 2: \quad 5n = 10 = n \cdot 5 \]
\[ \text{At } n = 2: \quad 5 - n = 3 \quad \text{but} \quad n - 5 = -3 \]
Why: Multiplication may be written in either order, so 5n and n times 5 are the same product for every value of n. Subtraction may not: at n equal to 2, five minus n is 3 while n minus five is negative 3. The two answers are opposite in sign, which is the clearest possible demonstration that the order in a subtraction is part of the meaning.
Explain it
They are comfortable with arithmetic and have never seen a letter used in place of a number.
Discussion prompt
In no more than four sentences, and without using the word variable, explain what 180t means and why anyone would write it that way instead of just doing the arithmetic. Then give them one test they can run on any expression to decide whether it is a multiplication.
Hint: The test is about what sits next to what, with nothing between them.
Answer:
A usable answer: 180t is a rule rather than an answer. It says take however many hours you have travelled and multiply it by 180, and it says that once for every possible number of hours instead of once for each one you happen to care about. That compression is the entire reason for the letter.
The test: if two things are written right next to each other with no sign between them — a number and a letter, two letters, or two brackets — they are being multiplied. It is the only operation algebra ever writes as nothing at all.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: Naming operations is fixed by reading four or five expressions aloud until the hidden signs stop being invisible. Evaluating is fixed by writing the substitution line separately every time, even when it feels unnecessary. Units are fixed by writing them alongside every number you substitute and seeing what survives. Writing your own expression is fixed by testing every one you write at two values, one of which is zero. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Divide a page into three horizontal strips. In the top strip, write the four operations and beside each one write every way algebra can express it — putting a star next to the two that can be written with no sign at all. In the middle strip, draw the three boxes of the evaluation routine, and under them work one example of your own choosing all the way through, showing all three lines. In the bottom strip, write the two formulas from this lesson, and beside each one write the unit its answer carries and where that unit came from. Finally, in the margin, write the single expression from this lesson that changes when you swap its two parts, and write its swapped version next to it.
The marginal pair should be a subtraction or a division, not an addition or a multiplication. If you wrote down an addition, go back to the comparison table.
Recap
Five things, and the fifth one is the one that keeps checking your work for the rest of the year.
| If the question says | Your first move is |
|---|---|
| Identify the variable | Look for the letters and ignore everything else |
| State the meaning and the operation | Check whether a sign is hidden between the symbols |
| Evaluate when x equals something | Write the expression, then the substitution, on separate lines |
| Use the formula to find | Write the formula in letters before any number goes in |
| Write an expression for | Decide which quantity varies, and name it with a letter that reminds you what it is |
Lesson 1.2 keeps the same routine and adds one new operation to it: raising a number to a power. Nothing about write, substitute, simplify changes — the expressions simply get one layer deeper.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 1 Connections to Algebra — Lesson 1.1 Variables in Algebra §1.1, pp. 3-8 — everything on these slides traces back here
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