Coefficients including the invisible ones, identifying like terms by matching the variable part exactly, combining them by adding coefficients as the distributive property run backwards, simplifying expressions that contain grouping symbols, and knowing when an expression counts as simplified.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 2 — Properties of Real Numbers
Combining Like Terms
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. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-112 — the lesson these objectives are drawn from
Warm-up
Combining like terms is counting, and you can already count.
Discussion prompt
You have 8 identical pens and someone gives you 3 more. How many pens do you have? Now do the same with 8x and 3x. What is the same about the two questions?
Hint: In both cases the thing being counted never changes.
Answer:
\[ 8x + 3x = 11x \]
Eleven pens and eleven x, by the same reasoning. The x is a thing being counted, and adding eight of them to three of them gives eleven of them. The only reason this feels harder in algebra is that you cannot see what x is — and you do not need to, because the count works whatever it turns out to be.
Concept
Two terms with the same variable part can be combined by adding their coefficients. That is not a new rule — it is the distributive property from Lesson 2.6 read from right to left, with the shared variable factored out.
like terms — Terms in an expression that have the same variable raised to the same power. Numbers on their own are also considered like terms.
\[ 8x + 3x = (8 + 3)x = 11x \]
The variable never changes during the combination. Only the count in front of it does.
Figure (svg): The distributive property run backwards, turning 8x plus 3x into the quantity 8 plus 3 all times x
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-108
Section
Section 1
Concept
In a term that is the product of a number and a variable, the number is called the coefficient of the variable. Two coefficients are written invisibly and have to be read in.
coefficient — The number multiplying the variable in a term. A variable written alone has a coefficient of one; a variable with a minus sign in front has a coefficient of negative one.
Figure (svg): The terms x and 3 x squared labelled with their coefficients, showing that x has an unwritten coefficient of 1
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107 — the definition of coefficient and the Reading Algebra note
Picture it
One of these coefficients is written and one is not.
Figure (svg): The terms x and 3 x squared labelled with their coefficients, showing that x has an unwritten coefficient of 1
The invisible coefficients are the ones that get dropped when terms are combined. Writing them in the first time you meet an expression costs nothing and prevents a whole family of errors.
Worked example
Two of these are the ones people miss.
\[ \text{Name the coefficient in each term: } \; 3x^2, \quad x, \quad -x, \quad -7y, \quad y^2. \]
Three x squared has coefficient 3
Why: The number multiplying the variable part is written explicitly.
\[ 3 \]
x has coefficient 1
Why: A variable alone means one of it, so the coefficient is one even though it is not written.
\[ 1 \]
The opposite of x has coefficient negative 1
Why: A minus sign in front of a bare variable is a coefficient of negative one, by the property of negative one from Lesson 2.5.
\[ -1 \]
Negative 7y has coefficient negative 7
Why: The sign belongs to the coefficient, not to the variable.
\[ -7 \]
y squared has coefficient 1
Why: Again a variable part alone, so the count is one.
\[ 1 \]
Figure (svg): The solution to Worked example name every coefficient shown as a ladder of expressions, one row per algebraic move
\[ 3, \quad 1, \quad -1, \quad -7, \quad 1 \]
Verify: rewrite each term with its coefficient shown
Why: Writing them as 3x squared, 1x, negative 1x, negative 7y and 1y squared reproduces the originals exactly, since multiplying by one changes nothing. That check confirms the invisible coefficients were read as one rather than as zero.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107
Matching
Two of these coefficients are not written down.
Match the pairs
Why: The middle two are the ones worth practising: a variable alone counts as one of itself, and a variable with a minus sign counts as negative one. Both facts come from the property of negative one and the multiplicative identity, and both are used every time like terms are combined.
Worked example
Writing the expression as a sum is what attaches each sign to its own term.
\[ \text{Name the coefficient of each term in } \; x^2 - 5x + 4 - 3x. \]
Write the expression as a sum
Why: Every subtraction becomes an addition of the opposite, by the rule from Lesson 2.4.
\[ x ^{2} + (-5 x) + 4 + (-3 x) \]
Read the first coefficient
Why: x squared alone has coefficient one.
\[ 1 \]
Read the next two
Why: Negative five x has coefficient negative five; the four is a constant with no variable.
\[ -5\text{ and } 4 \]
Read the last
Why: Negative three x has coefficient negative three.
\[ -3 \]
Figure (svg): The solution to Worked example the sign belongs to the coefficient shown as a ladder of expressions, one row per algebraic move
\[ 1, \quad -5, \quad -3 \quad \text{and the constant } 4 \]
Verify: reassemble the expression from the coefficients
Why: One x squared plus negative five x plus four plus negative three x rewrites as x squared minus five x plus four minus three x, which is the original. Nothing was lost or invented, which is what the rewrite is for.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107
Trap
\[ 2y^2 + 7y^2 - y^2 \]
Add the visible coefficients two and seven, and ignore the last term because it has no number
Why: The third term looks different from the other two, so it gets treated as something other than a count.
\[ = 9y^2 \quad \text{(wrong)} \]
The last term is negative one y squared, so it takes one away rather than contributing nothing. The correct total is eight.
\[ 2y^2 + 7y^2 - y^2 = 2y^2 + 7y^2 + (-1)y^2 \]
Write in the invisible coefficient before adding anything
Why: A bare variable part has a coefficient of one, and a minus sign in front makes it negative one.
\[ = (2 + 7 - 1)y^2 = 8y^2 \]
Reading the invisible coefficient is a habit worth ten seconds. It is the single most common slip in this lesson, and it is invisible in the answer.
Sorting
Write each expression as a sum first, then read the sign of the named term.
Sort into buckets
Sort each term by the sign of its coefficient.
The two bare-variable cases are the ones to watch. A coefficient of one and a coefficient of negative one are both invisible on the page and both fully present in the arithmetic.
Faded example
Supply the coefficients that are not written.
Fill in the blanks
x^2 + 5x - x = 1x^2 + 5x + (-1)x
Why: A bare x squared means one of it, and the minus x means negative one of it. Writing both coefficients in makes the expression's arithmetic visible: the x terms are now five and negative one, which combine to four. Nothing about the expression changed — only what is written down.
Socratic
The invisible coefficient is not a convention someone chose.
Discussion prompt
Explain, using a property from Lesson 2.5, why writing x is the same as writing one times x. Then say why that matters when combining like terms.
Hint: One of the six properties of multiplication is exactly about this.
Answer:
The identity property of multiplication says one times a equals a for every number a. So one x and x name the same quantity, and writing the one is optional rather than wrong.
It matters because combining like terms means adding coefficients, and a term with no visible coefficient still has one. Treating x as having no coefficient would mean adding nothing when the term should contribute one — and the same reasoning gives negative one for a bare variable behind a minus sign, via the property of negative one.
Section
Section 2
Concept
Like terms have the same variable raised to the same power. Numbers on their own are considered like terms. The coefficients play no part in deciding — only the variable part does.
Writing the expression as a sum first is what makes the terms and their signs visible.
Figure (svg): An expression with its like terms grouped by colour, showing x squared alone, two x terms and two constants
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107 — the definition of like terms and Example 1
Picture it
Each colour marks a set of like terms.
Figure (svg): An expression with its like terms grouped by colour, showing x squared alone, two x terms and two constants
The x squared term is alone in its group, which is perfectly normal — a term with no partner simply survives the simplification unchanged.
Worked example
This is Example 1 from the textbook. Rewrite as a sum before deciding anything.
\[ \text{Identify the like terms in } \; x^2 - 5x + 4 - 3x + 2. \]
Write the expression as a sum
Why: Each subtraction becomes an addition of the opposite, so every term carries its sign.
\[ x ^{2} + 5 x + (-4) + (-3 x) + 2 \]
Group the terms by variable part
Why: One x squared term, two x terms, and two constants.
Name the x terms
Why: Five x and negative three x share the variable part x.
\[ 5 x\text{ and } -3 x \]
Name the constants
Why: Negative four and two are both numbers with no variable.
\[ -4\text{ and } 2 \]
Figure (svg): An expression with its like terms grouped by colour, showing x squared alone, two x terms and two constants
\[ 5x \text{ and } -3x; \quad -4 \text{ and } 2 \]
Verify: check that the x squared term has no partner
Why: There is only one term with x squared in it, so it is in a group by itself. Counting the groups against the number of distinct variable parts confirms nothing was missed: x squared, x, and constant makes three.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107
Sorting
Compare the variable parts only, ignoring the coefficients.
Sort into buckets
Sort each pair by whether they are like terms.
The pair with matching coefficients but different letters is the trap. Coefficients are what get added during a combination, and they never decide whether a combination is allowed.
Worked example
Guided Practice 1 and 2. The second has two terms with the same power.
\[ \text{Identify the like terms in } \; 5x^2 + x - 8 - 6x + 10 \; \text{ and } \; 3x^2 - 2x + x^2 - 4 + 7x. \]
Rewrite the first as a sum and group
Why: The x terms are x and negative six x; the constants are negative eight and ten.
\[ x\text{ with } -6 x, -8\text{ with } 10 \]
Note the lone term in the first
Why: Five x squared has no partner, so it stands alone.
\[ 5 x ^{2}\text{ alone} \]
Rewrite the second as a sum and group
Why: The x squared terms are three x squared and x squared; the x terms are negative two x and seven x.
Note the lone term in the second
Why: Negative four is the only constant.
\[ -4\text{ alone} \]
Figure (svg): The solution to Worked example two more from guided practice shown as a ladder of expressions, one row per algebraic move
\[ x, -6x; \; -8, 10 \qquad 3x^2, x^2; \; -2x, 7x \]
Verify: check the invisible coefficients
Why: In the first expression the term x has coefficient one, and in the second the term x squared has coefficient one. Both would contribute nothing if the coefficient were read as zero, so noticing them now is what makes the combination in the next section correct.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-107
Error analysis
The student identified like terms in two expressions. Two of the four claims are wrong.
Annotate
On: \( \begin{aligned} \text{In } x^2 + 3x: & \; x^2 \text{ and } 3x \text{ are like terms} \\ \text{In } 5x - 2x: & \; 5x \text{ and } -2x \text{ are like terms} \\ \text{In } 4x + 4y: & \; 4x \text{ and } 4y \text{ are like terms} \\ \text{In } -8 + 10: & \; -8 \text{ and } 10 \text{ are like terms} \end{aligned} \)
Both errors come from looking at the wrong part of the term. The test is entirely about the variable part — letter and power — and never about the coefficient.
Elimination
Only one of these pairs shares a variable part.
Eliminate the wrong options
Which pair can be combined into a single term?
Survives elimination: A
Why: Both terms have the variable part y squared, so they may be combined by adding coefficients, giving nine y squared. The three wrong options each fail the same test in a different way — different power, missing variable, and different letter — and none of them can be simplified at all.
Discrimination
Strip the coefficient off each term and compare what is left.
Sort into buckets
Sort each term by its variable part.
Counterexample
One case is enough to refute a general statement.
Discussion prompt
A student claims that two terms are alike whenever they contain the same letter. Give a counterexample, and then state the corrected version of the claim.
Hint: Look for two terms with the same letter that cannot be combined.
Answer:
\[ x^2 \text{ and } x \text{ both contain } x, \text{ but they are not like terms} \]
At x equal to three, x squared is nine and x is three, so they do not even represent the same kind of quantity. Combining them into four x or four x squared would give the wrong value at almost every input.
The corrected claim is that two terms are alike when they contain the same letter raised to the same power. The power is as much a part of the variable part as the letter is, which is why an area and a length can never be added — the same reasoning as the square feet and cubic feet of Lesson 1.2.
Section
Section 3
Concept
The distributive property allows like terms to be combined by adding their coefficients. Factoring the shared variable out of the terms turns the sum of terms into a single term.
\[ 8x + 3x = (8 + 3)x = 11x \]
The property extends to three or more terms in the same way, since the variable can be factored out of all of them at once.
Figure (svg): The distributive property run backwards, turning 8x plus 3x into the quantity 8 plus 3 all times x
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108 — the Simplified Expressions paragraph and Example 2
Picture it
The tile never changes; only how many there are.
Figure (svg): Algebra tiles showing eight x tiles plus three x tiles making eleven x tiles
Whatever x turns out to be, eleven of them is eight of them plus three of them. That is why the combination is valid without knowing the value of x at all.
Worked example
This is Example 2 from the textbook. The second has an invisible coefficient.
\[ \text{Simplify } \; 8x + 3x \; \text{ and } \; 2y^2 + 7y^2 - y^2 + 2. \]
Factor x out of the first
Why: The distributive property backwards: eight x plus three x is the quantity eight plus three, all times x.
\[ (8 + 3) x \]
Add the coefficients
Why: Eight and three make eleven.
\[ 11 x \]
Write in the invisible coefficient in the second
Why: The term minus y squared has coefficient negative one.
\[ 2 y ^{2} + 7 y ^{2} + (-1) y ^{2} + 2 \]
Factor and add
Why: Two plus seven minus one is eight, so the y squared terms combine to eight y squared, and the constant two is left alone.
\[ 8 y ^{2} + 2 \]
Figure (svg): The solution to Worked example two combinations shown as a ladder of expressions, one row per algebraic move
\[ 8x + 3x = 11x \qquad 2y^2 + 7y^2 - y^2 + 2 = 8y^2 + 2 \]
Verify: substitute a value into the second
Why: At y equal to 1 the original is two plus seven minus one plus two, which is ten, and the simplified form is eight plus two, also ten. The two agree, which they must if the combination was legal.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108
Faded example
The variable has been factored out. Supply the arithmetic.
Fill in the blanks
2y^2 + 7y^2 - y^2 = (2 + 7 - 1)y^2 = 8y^2
Why: The third term has an invisible coefficient of negative one, so the arithmetic inside the bracket is two plus seven minus one, which is eight. Factoring the y squared out first is what reduces the whole problem to a single line of ordinary addition.
Worked example
Guided Practice 3 to 5. The last one has terms that cancel entirely.
\[ \text{Simplify } \; 5x + 2x, \quad 8m - m + 3m - 5, \quad x^2 + 5x - x^2. \]
Combine the first
Why: Five plus two is seven.
\[ 7 x \]
Write in the invisible coefficient in the second
Why: The term minus m has coefficient negative one, so the m terms are 8, -1 and 3.
\[ 8 - 1 + 3 = 10 \]
Write the second answer
Why: Ten m, with the constant negative five left alone since it has no partner.
\[ 10 m - 5 \]
Combine the third
Why: The x squared terms have coefficients one and negative one, which total zero, so they vanish entirely.
\[ 5 x \]
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ 7x, \quad 10m - 5, \quad 5x \]
Verify: substitute into the third
Why: At x equal to 2 the original is four plus ten minus four, which is ten, and five x is also ten. The x squared terms really did cancel, which is what a coefficient total of zero means — the term disappears rather than becoming zero x squared written down.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108
Trap
\[ 8x + 3x \]
Add the coefficients and also combine the variables, giving eleven x squared
Why: Adding is happening, so it feels as though everything in the terms should be added or multiplied together.
\[ = 11x^2 \quad \text{(wrong)} \]
At x equal to two the original is sixteen plus six, which is twenty-two, and eleven x squared is forty-four. The variable part must survive unchanged.
\[ 8x + 3x = (8 + 3)x = 11x \]
Factor the variable out and add only what is left
Why: The variable is a common factor, and factoring it out leaves only the coefficients to be added.
The tile picture makes it obvious: eight tiles and three tiles make eleven tiles, and the tiles do not change shape when you count them.
Elimination
The expression is 8m minus m plus 3m minus 5.
Eliminate the wrong options
Which is correct?
Survives elimination: A
Why: The m coefficients are eight, negative one and three, totalling ten. The constant negative five has no like partner, so it survives unchanged. Substituting m equal to one confirms it: the original is eight minus one plus three minus five, which is five, and ten minus five is also five.
Prediction
Sometimes a whole group of like terms disappears.
Predict first
What does x squared plus 5x minus x squared simplify to?
Correct: 5x.
\[ x^2 + 5x - x^2 = (1 - 1)x^2 + 5x = 0 + 5x = 5x \]
Why: The x squared coefficients are one and negative one, which total zero, so zero times x squared is zero and that term vanishes entirely. The five x has no partner and survives. Substituting x equal to two gives four plus ten minus four, which is ten, and five x is also ten.
Socratic
You do not know what x is, and you are still allowed to combine.
Discussion prompt
Explain why 8x plus 3x can be simplified to 11x without knowing the value of x, and say why 8x plus 3y cannot be simplified at all.
Hint: Think about what the distributive property lets you factor out.
Answer:
The distributive property says the quantity eight plus three, all times x, equals eight x plus three x, for every value of x. So the two expressions are equal whatever x is, and replacing one by the other is safe without knowing anything about x. The shared factor is what makes it possible.
Eight x plus three y has no shared variable factor, so there is nothing to factor out and nothing to add. You could factor out a common numerical factor if there were one, but the letters cannot be merged — and substituting different values for x and y shows immediately that no single term could equal the sum for every pair.
Section
Section 4
Concept
When an expression contains brackets, the brackets have to go before the like terms can be seen. Distribute first, then group the like terms together, then combine them.
The order matters: grouping before distributing hides terms that are still inside brackets.
Figure (svg): A four-line simplification of 8 minus 2 times the quantity x plus 4, ending at the opposite of 2x
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108 — Example 3, Simplify Expressions with Grouping Symbols
Picture it
Five lines, and the annotation on each says which move was made.
Figure (svg): A four-line simplification of 8 minus 2 times the quantity x plus 4, ending at the opposite of 2x
Notice that the constants cancelled completely, leaving a single term. Writing one move per line is what makes such a cancellation visible rather than surprising.
Worked example
This is Example 3 part a from the textbook.
\[ \text{Simplify } \; 8 - 2(x + 4). \]
Distribute the negative 2
Why: The factor is negative two, sign included, and it reaches both terms.
\[ 8 + (-2) (x) + (-2) (4) \]
Multiply
Why: Negative two x and negative eight.
\[ 8 - 2 x - 8 \]
Group the like terms
Why: The constants are eight and negative eight; the x term stands alone.
\[ -2 x + (8 - 8) \]
Combine
Why: Eight and negative eight total zero, so only the x term survives.
\[ -2 x \]
Figure (svg): A four-line simplification of 8 minus 2 times the quantity x plus 4, ending at the opposite of 2x
\[ 8 - 2(x + 4) = -2x \]
Verify: substitute a value
Why: At x equal to 3 the original is eight minus two times seven, which is eight minus fourteen, or negative six. And negative two times three is also negative six. The constants really did cancel.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108
Ranking
Four moves, one correct sequence.
Put in order
Why: Distributing has to come first, because a term inside a bracket is not yet a term of the expression. Grouping comes second and requires the expression to be a sum so the terms may be reordered. Combining is third, and the substitution check is last since it needs a finished answer to compare against.
Worked example
Example 3 part b. Distribute both before grouping anything.
\[ \text{Simplify } \; 2(x - 3) + 3(5 - x). \]
Distribute the 2 over the first bracket
Why: Two x and negative six.
\[ 2(x) - 2(3) \]
Distribute the 3 over the second bracket
Why: Fifteen and negative three x.
\[ 3(5) - 3(x) \]
Group the like terms
Why: The x terms are two x and negative three x; the constants are negative six and fifteen.
\[ 2 x - 3 x - 6 + 15 \]
Combine each group
Why: Two minus three is negative one, giving the opposite of x; negative six plus fifteen is nine.
\[ -x + 9 \]
Figure (svg): The solution to Worked example two brackets shown as a ladder of expressions, one row per algebraic move
\[ 2(x - 3) + 3(5 - x) = -x + 9 \]
Verify: substitute a value
Why: At x equal to 2 the original is two times negative one plus three times three, which is negative two plus nine, or seven. And the opposite of two plus nine is also seven. Both brackets were distributed and both groups combined correctly.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108
Trap
\[ 8 - 2(x + 4) \]
Combine the 8 with the 4 inside the bracket, since both are constants
Why: They are both numbers with no variable, so they look like like terms.
\[ = -2(x + 12) \;? \text{ or } \; 12 - 2x \quad \text{(wrong)} \]
The four is inside a bracket that has not been distributed yet, so it is not yet a term of the outer expression. It has to be multiplied by negative two before it can join anything.
\[ 8 - 2(x + 4) = 8 - 2x - 8 = -2x \]
Distribute every bracket before identifying any like terms
Why: A number inside a bracket is not a term of the outer expression until the bracket is gone.
A quick check at x equal to three: the true value is negative six, while twelve minus two x would give six. Substitution catches this as reliably as it catches a missed distribution.
Elimination
The expression is 9x minus 4 times the quantity 2x plus 1.
Eliminate the wrong options
Which is correct?
Survives elimination: A
Why: Distributing negative four gives negative eight x and negative four, so the expression becomes nine x minus eight x minus four, which combines to x minus four. Substituting x equal to one confirms it: the original is nine minus twelve, which is negative three, and one minus four is also negative three.
Faded example
Both brackets are distributed. Group and combine.
Fill in the blanks
2x - 6 + 15 - 3x = (2 - 3)x + (-6 + 15) = -x + 9
Why: The x coefficients are two and negative three, totalling negative one, and the constants are negative six and fifteen, totalling nine. Writing each group's arithmetic in its own bracket before evaluating is what keeps the two groups from being mixed together.
Socratic
The order of the two moves is not a matter of taste.
Discussion prompt
Explain what goes wrong if you try to group like terms before removing the brackets, using 8 minus 2 times the quantity x plus 4 as your example. Then say what the brackets are doing to the terms inside them.
Hint: Ask whether the four inside the bracket is really a constant term of the whole expression.
Answer:
Grouping first tempts you to pair the eight with the four, since both are numbers. But the four is not a term of the outer expression — it is inside a bracket that is about to be multiplied by negative two, and after that multiplication it becomes negative eight rather than four.
A bracket suspends its contents from the surrounding expression until the multiplication is carried out. Only once the bracket is gone do the things inside become terms that can be grouped with anything outside. That is the same principle as the innermost-first rule from Lesson 1.3, applied to algebra rather than to arithmetic.
Section
Section 5
Concept
An expression is simplified if it has no grouping symbols and all its like terms have been combined. Both conditions have to hold — an expression with brackets removed but terms uncombined is not finished.
A term with no like partner is already as combined as it can be, and it survives into the answer unchanged.
Figure (svg): An expression rewritten as a sum so that every term carries its own sign before combining
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108 — the definition of a simplified expression
Picture it
Every term takes its sign with it when it moves.
Figure (svg): An expression rewritten as a sum so that every term carries its own sign before combining
The rewrite is what makes the reordering legal, exactly as in Lesson 2.4. Moving a term without its sign is the way grouping most often goes wrong.
Worked example
Both conditions have to be checked, and each one fails somewhere below.
\[ \text{Which of these are simplified? } \; 3x + 5, \quad 2(x + 1), \quad 4x + 2x, \quad x^2 + 3x. \]
Check the first
Why: No brackets, and three x and five are not alike, so it is simplified.
Check the second
Why: It contains a bracket, so it fails the first condition regardless of anything else.
Check the third
Why: No brackets, but four x and two x are alike and have not been combined.
Check the fourth
Why: No brackets, and x squared and x are not alike, so nothing can be combined.
Figure (svg): The solution to Worked example decide whether each is simplified shown as a ladder of expressions, one row per algebraic move
\[ 3x + 5 \text{ and } x^2 + 3x \text{ are simplified} \]
Verify: simplify the two that failed
Why: The second becomes two x plus two and the third becomes six x, and both of those pass the two conditions. An expression that fails a condition can always be simplified further, which is what makes the two conditions a usable test.
Sorting
Check both conditions on each expression.
Sort into buckets
Sort each expression by whether it is fully simplified.
Three of the six are finished and three are not, and the reasons are split between the two conditions. Checking both, rather than just looking for brackets, is what makes the test complete.
Worked example
Guided Practice 6 to 8. Each has a bracket to remove first.
\[ \text{Simplify } \; 3(y + 2) - 4y, \quad 9x - 4(2x + 1), \quad -(z + 2) + 2(1 - z). \]
Distribute and combine the first
Why: Three y plus six minus four y gives negative y plus six.
\[ -y + 6 \]
Distribute and combine the second
Why: Nine x minus eight x minus four gives x minus four.
\[ x - 4 \]
Distribute the leading minus in the third
Why: A minus sign in front of a bracket is a factor of negative one, so it gives negative z minus two.
\[ -z - 2 \]
Distribute the second bracket and combine
Why: Two minus two z, so the z terms are negative one and negative two, totalling negative three, and the constants are negative two and two, totalling zero.
\[ -3 z \]
Figure (svg): The solution to Worked example three from guided practice shown as a ladder of expressions, one row per algebraic move
\[ -y + 6, \quad x - 4, \quad -3z \]
Verify: substitute into the third
Why: At z equal to 1 the original is the opposite of three plus two times zero, which is negative three. And negative three times one is also negative three. The leading minus sign was distributed rather than dropped.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 108-108
Trap
\[ -(z + 2) \]
Remove the bracket and keep the terms as they were, since there is no number to distribute
Why: There is no visible coefficient, so it looks as though nothing needs multiplying.
\[ = z + 2 \quad \text{or} \quad -z + 2 \quad \text{(both wrong)} \]
A minus sign in front of a bracket is a factor of negative one, and it reaches every term inside. The correct expansion is negative z minus two.
\[ -(z + 2) = -1(z + 2) = -z - 2 \]
Write the invisible negative one before distributing
Why: The property of negative one from Lesson 2.5 says the minus sign is a factor, and factors distribute.
A check at z equal to one: the original is the opposite of three, which is negative three, and negative one minus two is also negative three.
Elimination
The expression is the opposite of the quantity z plus 2.
Eliminate the wrong options
What does it expand to?
Survives elimination: A
Why: A minus sign in front of a bracket is a factor of negative one, and by the distributive property it reaches every term inside. Both terms therefore change sign, giving negative z minus two. Substituting z equal to one confirms it: the original is negative three, and so is the answer.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Expression | Condition it fails | Simplified form |
|---|---|---|
| 2(x + 1) | contains a grouping symbol | 2x + 2 |
| 4x + 2x | like terms not combined | 6x |
| 3x + 5 | fails neither | 3x + 5 |
The bottom row is the important one: an expression with two terms can be fully simplified. Being short is not the test — the test is whether either condition still fails.
Socratic
The unsimplified expression gives the same values. Simplifying still earns its place.
Discussion prompt
Give two reasons for simplifying an expression before using it, at least one of which is about work you will do later in the book. Use 8 minus 2 times the quantity x plus 4 as your example.
Hint: Think about evaluating it at ten different values.
Answer:
First, evaluation becomes much cheaper. Evaluating the original at ten values means ten distributions and ten subtractions; evaluating the opposite of two x means ten single multiplications. The simplified form is the same function with far less work per input.
Second, the structure becomes visible. The simplified form shows immediately that the constants cancelled and the expression is a plain multiple of x, which the original hides completely. In Chapter 3 that structure is what tells you how to solve an equation, and in Chapter 4 it is what tells you what the graph looks like — neither is readable until the expression is simplified.
Comparison
Fill the blanks from memory before you scroll back. The test never looks at the coefficients.
Comparison matrix
| Pair | Alike? | Reason |
|---|---|---|
| 8x and -3x | Yes | same letter, same power |
| x squared and x | No | same letter, different power |
| 3x and 3y | No | different letters |
| -4 and 2 | Yes | both are numbers with no variable |
The third row is the trap: the coefficients match exactly and the terms are still not alike. Coefficients are what get added, never what decides whether adding is allowed.
Pattern
Whether the expression has brackets or not, the same five moves cover it.
Step four is where the invisible coefficients matter. A bare variable contributes one and a bare variable behind a minus sign contributes negative one, and neither is written down for you.
OpenStax Elementary Algebra 2e, §1.9 Properties of Real Numbers §1.9
Check
Invisible coefficients. Write them in first.
Check your understanding
Simplify 6y minus y plus 2y.
Answer: A
Why: The coefficients are six, negative one and two, which total seven. The middle term is a bare variable behind a minus sign, so its coefficient is negative one rather than nothing.
Check
Brackets first. Distribute before grouping.
Check your understanding
Simplify 5 minus 3 times the quantity x minus 2.
Answer: A
Why: Distributing negative three gives negative three x plus six, so the expression is five minus three x plus six, which combines to negative three x plus eleven. Substituting x equal to one confirms it: the original is five minus three times negative one, which is eight, and the answer gives negative three plus eleven, also eight.
Check
Like terms. Compare the variable parts only.
Check your understanding
Which expression is already fully simplified?
Answer: A
Why: There are no grouping symbols, and x squared and three x are not like terms because the powers differ. Both conditions hold, so nothing further can be done — and having two terms does not stop an expression from being simplified.
Real world
You are pricing a party. Each guest costs 12 dollars for food and 5 dollars for a favour, the hall costs 200 dollars, and a discount of 3 dollars per guest applies.
Discussion prompt
Write an expression for the total cost with n guests, using one term for each item described, then simplify it. Say what the simplified form tells you at a glance that the original does not, and use it to find the cost for 30 guests.
Hint: Three of the four items depend on n and one does not.
Answer:
\[ 12n + 5n + 200 - 3n \]
\[ = (12 + 5 - 3)n + 200 = 14n + 200 \]
The simplified form says immediately that each additional guest costs fourteen dollars and that two hundred is fixed whatever happens. The original expression contains that information but hides it across four terms, and nobody planning a party could read it off.
\[ \text{at } n = 30: \quad 14(30) + 200 = 420 + 200 = 620 \text{ dollars} \]
This fixed-plus-per-unit shape is the one from Lesson 1.1, and combining like terms is what recovers it from a messy description. Chapter 4 will call the fourteen the rate of change and the two hundred the starting value.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Can 3x and 3y be combined into a single term?
Correct: No, the variable parts differ.
\[ \text{at } x = 1, y = 2: \quad 3x + 3y = 9 \quad \text{but} \quad 6xy = 12 \]
Three x plus three y can be written as three times the quantity x plus y, which factors out the common coefficient — but that is not a single term.
Why: Terms are alike when their variable parts match exactly, and here one is x and the other is y. The matching coefficients are irrelevant — coefficients are what get added once a combination is allowed, never what permits it. Substituting x equal to 1 and y equal to 2 gives three plus six, which is nine, while six x y would give twelve and six x would give six. Neither proposed combination works.
Explain it
They can distribute and have never combined like terms.
Discussion prompt
In no more than four sentences, explain what makes two terms combinable, using a counting analogy rather than a rule. Then give them the one thing they must check that is not written on the page.
Hint: The unwritten thing is a number.
Answer:
A usable answer: think of each term as a count of something. Eight x means eight of the thing called x, and three x means three more of the same thing, so together you have eleven of it. You can only add counts of the same thing — eight apples and three oranges is not eleven of anything.
The unwritten thing to check is the coefficient of a bare variable. A term written as just x means one of it, and a term written as minus x means negative one of it. Both numbers are fully present in the arithmetic and neither appears on the page, which is why they are the ones most often dropped.
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: Invisible coefficients are fixed by writing them in every time, before any arithmetic. Deciding likeness is fixed by covering the coefficients and comparing only what is left. Order of operations is fixed by remembering that a term inside a bracket is not yet a term of the expression. A bare minus sign is fixed by writing it as negative one before distributing. 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
At the top of a page write one expression with at least five terms including a bare variable, a negated variable and two constants. Underneath, rewrite it as a sum with every invisible coefficient written in, and draw a loop round each group of like terms in a different colour. To the right, show each group combining as a distributive-property step with the variable factored out. In the lower half, work one full simplification of an expression containing two brackets, writing one move per line and annotating each line with which move it was. Finally, in the margin, write the two conditions for an expression to be simplified and give one expression that fails each.
Every loop you drew should contain terms whose variable parts are identical. If any loop contains two different powers of the same letter, look again at the definition of like terms.
Recap
Five things, and the first one is the one that is invisible on the page.
| If the question says | Your first move is |
|---|---|
| Identify the like terms | Rewrite as a sum, then compare variable parts |
| Simplify the expression | Look for brackets and distribute them first |
| Combine 8m - m + 3m | Write the coefficient of -m as negative one |
| Simplify -(z + 2) | Write the minus sign as a factor of negative one |
| Is this simplified | Check both conditions, not just the brackets |
Lesson 2.8 finishes the chapter with division, using the same trick that made subtraction disappear: dividing by a number is multiplying by its reciprocal, so the multiplication rules you already have cover division too.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 2 Properties of Real Numbers — Lesson 2.7 Combining Like Terms §2.7, pp. 107-112 — everything on these slides traces back here
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