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Collect like terms, including fraction and decimal coefficients, so a long linear expression or equation becomes short enough to solve and check.
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By the end of this lesson you will be able to tidy a long algebraic expression by collecting like terms, including terms with fraction and decimal coefficients, and to explain why that is allowed. You will use collecting to shorten each side of a linear equation before you solve it, and you will check your work by substituting a value into the long and the short forms.
You can add and subtract negative numbers, fractions and decimals. You know that an expression like $4n$ means $4$ times $n$, that $n$ on its own means $1n$, and that $-n$ means $-1n$. You can expand a bracket such as $3(x - 2)$ and solve equations such as $5x - 4 = 2x + 11$ by doing the same thing to both sides. You also know that $4$ apples plus $3$ apples is $7$ apples, but $4$ apples plus $3$ pears is not $7$ of anything. That sentence is the whole idea of this lesson. Here we make it exact, extend it to fraction and decimal coefficients, and use it to turn long equations into short ones.
| Term | What it means |
|---|---|
| Term | One part of an expression, separated from the others by $+$ or $-$. The sign in front belongs to the term. |
| Coefficient | The number multiplying the letters in a term: $-3$ in $-3x$, $\frac{1}{2}$ in $\frac{1}{2}y$, $1$ in $x$. |
| Constant | A term with no letter, such as $7$ or $-2.5$. |
| Like terms | Terms with exactly the same letters raised to the same powers, such as $4x$ and $-9x$. |
| Collect like terms | Replace like terms by one term whose coefficient is the sum of theirs. |
| Equivalent expressions | Expressions that give the same value for every value of the letters. |
An expression such as $6m - 4 + 2m + 9 - m$ is a sum of terms. Each term has a coefficient and a letter part: $6m$, $2m$ and $-m$ all have letter part $m$, and $-4$ and $9$ have no letter at all. Terms with the same letter part are like terms, and like terms can be added, because they count the same thing. Six $m$'s plus two $m$'s minus one $m$ is seven $m$'s.
Collecting like terms means doing exactly that: add the coefficients of each group and keep the letter part. The expression becomes $7m + 5$. It looks different but it is equivalent: it gives the same value for every value of $m$. Try $m = 10$: the long form gives $60 - 4 + 20 + 9 - 10 = 75$, and the short form gives $70 + 5 = 75$.
Why is this allowed? The distributive property says $6m + 2m = (6 + 2)m$. That is the same rule you use to expand brackets, read backwards. And because addition can be done in any order, terms may be moved around, as long as each one takes its sign with it.
Collecting is the step that makes long equations solvable. Before you can undo and divide, each side of an equation should have at most one $x$ term and one number. Collecting gets it there.
Another way: algebra tiles
Lay out six long tiles for $6m$, two more for $2m$, and remove one for $-m$; add nine small unit tiles and take away four. Seven long tiles and five small ones are left: $7m + 5$.
Another way: a shopping list
"$3$ notebooks, $2$ pens, $1$ more notebook, $4$ more pens" becomes "$4$ notebooks, $6$ pens." Writing $n$ for notebook and $p$ for pen, $3n + 2p + n + 4p = 4n + 6p$.
Two terms are like when their letter parts match exactly: the same letters, each to the same power. The coefficients do not matter.
| Terms | Like? | Why |
|---|---|---|
| $5x$ and $-2x$ | yes | both are a number times $x$ |
| $5x$ and $5y$ | no | different letters |
| $3x$ and $3x^2$ | no | $x^2$ is $x \cdot x$, a different kind |
| $4xy$ and $-yx$ | yes | $yx$ is the same product as $xy$ |
| $7$ and $-0.5$ | yes | both are constants |
A good test is to picture what each term counts. $3x^2$ could be the area of three squares of side $x$; $3x$ is the length of three sides. Areas and lengths cannot be added into one kind of thing, so the terms stay apart. When nothing is like anything else, as in $2a + 5b - 3$, the expression is already collected.
The most common error in collecting is losing a minus sign. In $8y - 3 + 2y - 5y$, the terms are $8y$, $-3$, $+2y$ and $-5y$. Circle each term together with the sign in front of it. Then the $y$ group is $8 + 2 - 5 = 5$, and the collected form is $5y - 3$.
Think of subtracting as adding the opposite: $8y - 3 + 2y - 5y$ is $8y + (-3) + 2y + (-5y)$. Written that way, the terms can be put in any order, and the signs cannot get lost. A term at the start with no sign written is positive.
In grade 8, coefficients are often fractions or decimals, because real rates are. The rule does not change: add the coefficients.
With decimals, line them up: $0.4n + 1.25n = 1.65n$. With fractions, use a common denominator: $\frac{1}{2}a - \frac{1}{4}a = \frac{2}{4}a - \frac{1}{4}a = \frac{1}{4}a$. A letter with no number in front counts as $1$, so $b + \frac{2}{3}b = \frac{3}{3}b + \frac{2}{3}b = \frac{5}{3}b$.
When a coefficient comes out as $1$ or $-1$, write just $x$ or $-x$. When it comes out as $0$, the term disappears: $3x + 5 - 3x = 5$.
In an equation, collect each side on its own first. Do not move terms across the equals sign while you are collecting; that is a separate step.
Take $3x + 7 - x + 2 = 25$. The left side collects to $2x + 9$, so the equation is $2x + 9 = 25$. Now it is a two-step equation: $2x = 16$, $x = 8$. Check in the original equation: $24 + 7 - 8 + 2 = 25$.
If both sides have $x$ terms after collecting, you then gather them on one side, as you did with variables on both sides. Collecting first keeps the number of moves small, and fewer moves means fewer chances to slip.
Brackets often hide like terms. In $5(y - 2) + 3(2y + 1)$, nothing can be collected until the brackets are gone. Expand each one: $5y - 10$ and $6y + 3$. Now the expression is $5y - 10 + 6y + 3$, and the like terms are in plain sight: $11y - 7$.
A minus in front of a bracket needs extra care, because it changes the sign of every term inside: $8y - (3y - 4)$ is $8y - 3y + 4$, which collects to $5y + 4$. A common slip is to write $8y - 3y - 4$, which gives $5y - 4$. Checking with $y = 2$ catches it at once: $16 - 2 = 14$, and $5 \times 2 + 4 = 14$, but $5 \times 2 - 4 = 6$.
So the order of work is always the same: expand first, then collect, then solve if there is an equation.
Two expressions are equivalent when they agree for every value of the letter, so a quick check is to try one or two values. Pick values that are easy to work with but not $0$ or $1$, because those can hide mistakes: with $x = 0$, every $x$ term vanishes and a wrong coefficient goes unnoticed.
For $6m - 4 + 2m + 9 - m = 7m + 5$, try $m = 2$: the left side is $12 - 4 + 4 + 9 - 2 = 19$, and the right side is $14 + 5 = 19$. They agree. If they had not, one of the coefficients or signs would be wrong, and you would go back to the circled terms.
Why each move is allowed. Reordering uses the fact that addition can be done in any order; combining uses the distributive property, $ax + bx = (a + b)x$.
How to check. Substitute the same value, such as $2$ or $3$, into the original and the collected form. They must agree.
A family is fencing a rectangular backyard whose length is $12$ feet more than twice its width $w$. Walking around it, the four sides are $w$, $2w + 12$, $w$ and $2w + 12$. The perimeter is $w + 2w + 12 + w + 2w + 12$, which collects to $6w + 24$ feet. If the width is $30$ feet, the fence is $6 \times 30 + 24 = 204$ feet. At $18$ dollars a foot for a wood fence, that is $3{,}672$ dollars. The collected form makes it easy to try other widths before choosing a design.
A player's points come from two-point shots, three-point shots and free throws. Over three games she makes $4$, $6$ and $5$ two-pointers, $2$, $1$ and $3$ three-pointers, and $3$, $2$ and $4$ free throws. Her points are $(8 + 6 + 3) + (12 + 3 + 2) + (10 + 9 + 4)$. Collecting by kind of shot is tidier: $15$ two-pointers, $6$ three-pointers and $9$ free throws make $30 + 18 + 9 = 57$ points. The coach reads the collected form, not the game-by-game one, to see where the points come from.
A class plans a trip to a science museum for $s$ students. Each ticket is $18$ dollars and each boxed lunch $6$ dollars. The bus costs $450$ dollars, and a local business gives a $100$ dollar grant. The cost is $18s + 6s + 450 - 100$, which collects to $24s + 350$ dollars. For $40$ students that is $960 + 350 = 1{,}310$ dollars, or about $33$ dollars each. The collected form shows at a glance that each extra student adds $24$ dollars.
Leaving the sign behind. In $7a - 2b + 3a$, the term is $-2b$.
Combining unlike terms. $3x + 4$ is not $7x$, and $x^2 + x$ is not $2x^2$ or $x^3$.
Forgetting the invisible 1. $x + 5x = 6x$, not $5x$.
Multiplying the letters when adding. $2x + 3x = 5x$, not $5x^2$.
Moving terms across the equals sign while collecting. Collect each side first; then solve.
Collect $6m - 4 + 2m + 9 - m$. Circle the $m$ terms with their signs.
$+6m, \; +2m, \; -m$
The sign in front of each term belongs to it, and $-m$ is $-1m$.
Add the coefficients of the $m$ terms.
$6 + 2 - 1 = 7$
Seven $m$'s in all.
Combine the constants.
$-4 + 9 = 5$
Start at $-4$ and go up $9$.
Write the collected expression.
$7m + 5$
The letter term first, then the constant.
Check with $m = 3$.
$18 - 4 + 6 + 9 - 3 = 26, \quad 21 + 5 = 26$
Both forms give $26$, so they are equivalent.
Collect $\frac{1}{2}a + \frac{2}{3}b - \frac{1}{4}a + b$. Group the $a$ terms.
$\frac{1}{2}a - \frac{1}{4}a$
$a$ and $b$ are different letters, so they form separate groups.
Rewrite over a common denominator.
$\frac{2}{4}a - \frac{1}{4}a$
Fractions add easily once the denominators match.
Subtract the numerators.
$\frac{1}{4}a$
Two quarters minus one quarter is one quarter.
Add the $b$ terms, with $b$ as $\frac{3}{3}b$.
$\frac{2}{3}b + \frac{3}{3}b = \frac{5}{3}b$
A letter on its own has coefficient $1$.
Write the collected expression.
$\frac{1}{4}a + \frac{5}{3}b$
An $a$ term and a $b$ term are unlike, so this is as short as it gets.
Check with $a = 4$ and $b = 3$.
$2 + 2 - 1 + 3 = 6, \quad 1 + 5 = 6$
Both forms give $6$.
Solve $0.6x + 4 + 0.9x - 1 = 2.5x - 0.5x - 5$. Collect the left $x$ terms.
$0.6x + 0.9x = 1.5x$
Add the decimal coefficients.
Collect the left constants.
$4 - 1 = 3$
The left side is now $1.5x + 3$.
Collect the right side.
$2.5x - 0.5x - 5 = 2x - 5$
Each side is collected on its own.
Subtract $1.5x$ from both sides.
$3 = 0.5x - 5$
Gather the $x$ terms on the side with more of them.
Add $5$ to both sides.
$8 = 0.5x$
Undo the constant next to the $x$ term.
Divide both sides by $0.5$.
$x = 16$
Dividing by $0.5$ is the same as doubling.
Check the left side of the original equation.
$9.6 + 4 + 14.4 - 1 = 27$
$0.6 \times 16 = 9.6$ and $0.9 \times 16 = 14.4$.
Check the right side of the original equation.
$40 - 8 - 5 = 27$
Both sides give $27$, so $x = 16$ is the solution.
Collect the $k$ terms.
$5k - 2k = 3k$
The minus belongs to $2k$.
Collect the $j$ terms.
$-3j + 7j = 4j$
Start at $-3$ and go up $7$.
Write the collected expression.
Check with $k = 2$ and $j = 3$.
Which term can be collected with $2x$ in $2x + 8x^2 + 6y + 7x$?
Complete the worked solution that collects $8x + 17 + 2x - 5$ and checks it with $x = 2$.
Add the coefficients of the $x$ terms.
$8 + 2 =$ p
That many $x$'s in all.
Combine the plain numbers.
$17 - 5 =$ q
The minus belongs to the last number.
Put $x = 2$ into the collected expression.
$2 \times \text{coefficient} + \text{number} =$ r
The original expression gives the same value at $x = 2$, which checks the collecting.
Collect like terms: $6x - 8 + x + 4$.
Answer:
Collect like terms: $2x + 2y + 3x + 3y$.
Answer:
Collect like terms: $0.2n + 8 + 1.6n + 3$.
Answer:
Solve $6x + 5 - 2x - 4 + x = 26$.
Collected: k $x + 1 = 26$, so $x =$ x
At a farm stand, a family sells $x$ baskets of peaches for $7$ dollars each and $x$ jars of jam for $8$ dollars each. They pay $34$ dollars to rent the stand and get back a $9$ dollar deposit on their jars. Write the day's balance in dollars, collected into simplest form.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Simplify by collecting like terms: $9x - 8 - 5x + 2 + 3x$.
Answer:
You can collect like terms and use collecting to solve a linear equation. Without looking: why can $3x$ and $-5x$ be collected, but $3x$ and $3x^2$ cannot?
19. Your turn: collect $5k - 3j - 2k + 7j - 4$., step 3
$3k + 4j - 4$
The constant $-4$ has nothing to combine with.
19. Your turn: collect $5k - 3j - 2k + 7j - 4$., step 4
$10 - 9 - 4 + 21 - 4 = 14, \quad 6 + 12 - 4 = 14$
Both forms agree.