Equivalent expressions: the distributive property and like terms
Two people write different expressions for the same count. How can you tell that they will always give the same number, without trying every number there is?
Where you are 12 in Expressions
When you reach for it
A garden center sets out plant pots in 3 rows. Every row has some pots of basil, the same number in each row, and 4 pots of mint. Call the number of basil pots in a row x. Kaia counts one row and multiplies by the number of rows:
Rafa counts all the basil, then all the mint:
They wrote different expressions, but they counted the same pots, so they must get the same total whatever x is. Two expressions like these are called equivalent: they give the same value for every number you put in place of the letter.
You reach for equivalent expressions whenever one form of an expression is more useful than another. A form with no brackets is quicker to add to something else. A form with a factor at the front shows the rows, groups or packs hiding in a total. A form with fewer terms is easier to evaluate. And when two people (or two methods) produce different expressions, the question “are these equivalent?” is how you find out whether both are right.
This lesson takes three days. Part 1 makes equivalent expressions by distributing, with an area picture that shows why it works. Part 2 runs the same move backwards and factors out a common factor. Part 3 combines like terms, and then asks how to test whether two expressions are equivalent, and what a test can and cannot prove.
The procedure
Part 1: distributing
Two expressions are equivalent when they have the same value for every value of the letter. Not for one value, not for most values: for every one. So 3(x + 4) and 3x + 12 are equivalent only if no number at all makes them different.
Trying a few values is a good start. Putting x = 0, x = 1 and x = 5 into each:
| x | 3(x + 4) | 3x + 12 |
|---|---|---|
| 0 | 3 \times 4 = 12 | 0 + 12 = 12 |
| 1 | 3 \times 5 = 15 | 3 + 12 = 15 |
| 5 | 3 \times 9 = 27 | 15 + 12 = 27 |
They agree every time. But three values are not every value, and there are endless numbers left to try. The reason they always agree is the distributive property, which you already use with numbers. To find 6 \times 23, split 23 into 20 + 3 and multiply each part:
A number multiplying a sum can multiply each part of the sum instead, and the answer is the same. In letters, for any numbers a, b and c,
The area picture shows why. Draw a rectangle 3 tall and x + 4 wide. Its area is 3(x + 4). A dashed line splits it into a piece 3 by x, with area 3x, and a piece 3 by 4, with area 12.
The two pieces fill the whole rectangle, so their areas add to the whole area: 3(x + 4) = 3x + 12. Nothing in the picture depends on how wide the x piece is drawn. Stretch it or squeeze it and the argument is the same, which is why the two expressions agree for every x and not just the three in the table.
Writing 3(x + 4) as 3x + 12 is called distributing the 3, or expanding the brackets. The number outside reaches every term inside. When a term inside already has a coefficient, as in 2(5y + 3), the outside number multiplies that coefficient: 2 \times 5y = (2 \times 5)y = 10y, by the associative property of multiplication, which lets you regroup a product. So 2(5y + 3) = 10y + 6.
The same works with subtraction inside the brackets. Take a rectangle 5 tall and x wide, and cut off a strip 2 wide. What is left is 5 tall and x - 2 wide, and its area is the whole 5x less the strip’s 10:
Part 2: factoring out a common factor
Distributing turns a product into a sum. Reading the same equation from right to left turns a sum back into a product, and that is called factoring. With numbers, 20 + 28 can be written as 4 \times 5 + 4 \times 7, because 4 is a factor of both, and then
Both sides are 48. The 4 has been factored out: written once at the front instead of once in each term.
With a letter it works the same way. Look at 18x + 24. The number 6 divides both 18 and 24, so 18x = 6 \times 3x and 24 = 6 \times 4, and
In the area picture, two pieces of area 18x and 24 that are both 6 tall sit side by side. Their widths are 3x and 4, so together they make one rectangle 6 by 3x + 4.
Any common factor gives an equivalent expression. 18x + 24 is also 2(9x + 12), and also 3(6x + 8), and distributing either one gives back 18x + 24. But in 2(9x + 12) the terms inside still share a factor, 3, so the factoring is not finished. Use the greatest common factor of the coefficients and constants, here 6, and nothing common is left inside: 3x and 4 share no factor but 1. That is the fully factored form, and it is the one meant when a question says “factor”.
You can always check a factoring by distributing it back. 6(3x + 4) = 18x + 24, which is where we started.
Part 3: like terms, and testing equivalence
Adding a number to itself several times is multiplying. y + y + y is three y’s, so
The same holds when the terms already have coefficients. Four b’s and three more b’s are seven b’s. The distributive property, read backwards, says so: the b is a common factor of both terms, so
Terms that can be combined this way are like terms: they have the same letter raised to the same power, or they are both plain numbers (see Terms, coefficients and constants). 4b and 3b are like terms; 4b and 3 are not, so 4b + 3 stays two terms. x^2 and x are not like terms either, because they are different numbers: when x = 3, x is 3 but x^2 is 9 (see exponents).
When the like terms are scattered, gather them first. The commutative property of addition lets you change the order of terms, and the associative property lets you group them:
Distributing, factoring out and combining like terms each use a property that holds for every number. So an expression made from another by these moves is equivalent to it, every time, with no need to try values.
Testing whether two expressions are equivalent. Sometimes you are handed two expressions and asked whether they are equivalent. Put the same value of the letter into both and compare.
If the two values differ, even once, the expressions are not equivalent. Equivalent means “the same for every value”, and one value where they differ shows that this fails. One disagreement settles it.
If the values agree, that is evidence and not proof. Try x^2 + 6 and 5x:
| x | x^2 + 6 | 5x |
|---|---|---|
| 2 | 4 + 6 = 10 | 10 |
| 3 | 9 + 6 = 15 | 15 |
| 0 | 0 + 6 = 6 | 0 |
They agree at x = 2 and at x = 3, and anyone who stopped there would think they were equivalent. At x = 0 they differ, so they are not. To prove two expressions equivalent, rewrite one into the other using the properties. To disprove it, find one value where they differ. Values like 0 and 1 are quick to try first, and trying more than one value catches more.
Worked examples
Distributing
Write each expression without brackets.
(a) 5(m + 7) \qquad (b) 6(2k + 3) \qquad (c) \displaystyle \frac{1}{2}(10p + 4)
SolutionIn each one the number outside multiplies every term inside, by the distributive property.
(a) 5 \times m = 5m and 5 \times 7 = 35, so 5(m + 7) = 5m + 35.
(b) The first term inside already has a coefficient. Regrouping the product (the associative property), 6 \times 2k = (6 \times 2)k = 12k. Then 6 \times 3 = 18, so
(c) Multiplying by \displaystyle \frac{1}{2} is taking half. Half of 10p is 5p and half of 4 is 2, so \displaystyle \frac{1}{2}(10p + 4) = 5p + 2.
Check (b) with one value, say k = 1: 6(2 + 3) = 30 and 12 + 18 = 30.
Bubble tea for a group
Sunniva buys 4 cups of bubble tea for her friends. Each cup costs t dollars for the tea, plus 0.75 dollars for pearls.
(a) Write the total cost as a product and as a sum, and say what each form counts.
(b) The next week she buys 6 cups, each at t dollars, and a coupon takes 0.5 dollars off each cup. Write the total without brackets.
Solution(a) One cup costs t + 0.75 dollars, and there are 4 cups, so the total is 4(t + 0.75). That form counts cup by cup. Distributing,
This form counts tea, then pearls: 4t dollars of tea and 3 dollars of pearls. Both forms give the same total for every price t.
(b) One cup costs t - 0.5 dollars after the coupon, so 6 cups cost
That is the full price of the tea, 6t, less the 3 dollars the coupon saves in total. Check with t = 4: 6 \times 3.5 = 21 and 24 - 3 = 21.
Factoring out the greatest common factor
Factor each expression fully.
(a) 15w + 25 \qquad (b) 24a + 16 \qquad (c) 9c + 9 \qquad (d) 14x + 21y + 7
SolutionIn each one, find the greatest common factor of the numbers, write each term as that factor times something, and factor it out.
(a) The factors of 15 are 1, 3, 5, 15 and the factors of 25 are 1, 5, 25. The greatest they share is 5. Then 15w = 5 \times 3w and 25 = 5 \times 5:
(b) The greatest common factor of 24 and 16 is 8, so 24a + 16 = 8(3a + 2). Factoring out 4 instead gives 4(6a + 4), which is equivalent, but 6a and 4 still share a 2. With the 8 out, 3a and 2 share nothing but 1.
(c) Both terms are multiples of 9, and 9 = 9 \times 1. When a whole term is factored out, a 1 is left in its place:
Without the 1, the bracket would be 9(c), which is only 9c.
(d) There are three terms. The greatest common factor of 14, 21 and 7 is 7, so
Check by distributing: 7 \times 2x + 7 \times 3y + 7 \times 1 = 14x + 21y + 7.
Pots at the garden center
Uri works at a garden center.
(a) He has 42 pots of lavender and 30 pots of thyme. He wants identical rows, each with the same mix of lavender and thyme and nothing left over, and as many rows as possible. Write 42 + 30 as a product that shows the rows.
(b) Uri and three friends order 8 pots of pansies at x dollars each, plus a delivery charge of 20 dollars, and split the bill equally among the 4 of them. Write the bill, factor it fully, and say what the factored form tells each friend.
Solution(a) The number of rows has to divide both 42 and 30, and the most rows is their greatest common factor. The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30, and the largest of these that also divides 42 is 6. So
There are 6 rows, each with 7 pots of lavender and 5 pots of thyme. Both sides are 72 pots.
(b) The pansies cost 8x dollars and the delivery 20, so the bill is 8x + 20 dollars. The greatest common factor of 8 and 20 is 4:
The factored form is 4 equal shares of 2x + 5 dollars: each friend pays for 2 pots and 5 dollars of the delivery. The expanded form, 8x + 20, shows what the bill is made of; the factored form shows each friend’s part. Check with x = 3: the bill is 24 + 20 = 44 dollars, a quarter of that is 11, and 2 \times 3 + 5 = 11.
Combining like terms
Write each expression with as few terms as possible.
(a) w + w + w + w \qquad (b) 7d + 2 + 3d + 5 \qquad (c) y^2 + 5y + 2y + 3y^2
Solution(a) Four w’s added together is four times w: w + w + w + w = 4w.
(b) Reorder the terms (commutative property of addition) and group the like ones (associative property):
The 7d + 3d = (7 + 3)d step is the distributive property. 10d and 7 are not like terms, so 10d + 7 is as short as it gets.
(c) y^2 and 3y^2 are like terms, and so are 5y and 2y. A lone y^2 has coefficient 1:
4y^2 and 7y are not like terms, since y^2 and y are different numbers, so they stay apart.
Counting birds along a trail
Kaia walks a trail with 4 stops on a bird count. At every stop she counts the same number of gulls, g. At the first stop and at the last she also counts 3 herons, and she sees no herons at the two middle stops.
(a) Write her tally as a sum of the counts at each stop, then combine like terms.
(b) Rafa thinks of the trail as two halves, each with 2 stops. He writes 2(2g + 3). Show that his expression is equivalent to Kaia’s, and say what it counts.
Solution(a) Stop by stop the counts are g + 3, then g, then g, then g + 3. Added up and gathered:
(b) Distribute Rafa’s 2:
That is Kaia’s expression, so the two are equivalent: rewriting by the distributive property changes the form and never the value. Rafa’s form counts one half of the trail, 2g gulls and 3 herons, and doubles it. Notice that 4g + 6 factored fully is exactly 2(2g + 3), since the greatest common factor of 4 and 6 is 2.
Check with g = 5: Kaia gets 20 + 6 = 26 and Rafa gets 2 \times 13 = 26.
Equivalent or not?
Decide whether each pair is equivalent. Prove each answer.
(a) 2(x + 5) and 2x + 5 \qquad (b) 3m + 4 and 4m + 3 \qquad (c) 4(2n + 1) + 3n and 11n + 4
Solution(a) Try x = 0: 2(0 + 5) = 10 but 2 \times 0 + 5 = 5. They differ, so they are not equivalent. The 2 outside multiplies the 5 as well, so 2(x + 5) is 2x + 10.
(b) Try m = 1: 3 + 4 = 7 and 4 + 3 = 7. They agree, but that proves nothing yet. Try m = 0: 3m + 4 = 4 but 4m + 3 = 3. They differ there, so they are not equivalent.
(c) Rewrite the first expression. Distribute, then combine the like terms 8n and 3n:
The first expression has been turned into the second using only properties that hold for every number, so they are equivalent. A test at n = 2 agrees, as it must: 4 \times 5 + 6 = 26 and 22 + 4 = 26.
Boundaries
Equivalent means every value. Two expressions that agree at some values and not at others are not equivalent: x^2 + 6 and 5x agree at 2 and 3 only. Agreement at the values you tried is evidence; a rewrite by the properties is proof; one value where they differ is disproof.
The outside factor multiplies every term in the brackets. 3(x + 4) is 3x + 12. The distributive property is about multiplying a sum (or a difference). It does not spread a factor through a product: 2(5y) is 10y, a single term, because there is only one thing inside to multiply.
Only like terms combine. 6m + 5 is already as short as it can be. x and x^2 are not like terms, and neither are x and y.
A factor can be taken out only if it divides every term. In 8x + 5 the factors of 5 are 1 and 5, and 5 does not divide 8, so 8x + 5 has no common factor but 1. Any common factor gives an equivalent expression; only the greatest common factor leaves the bracket with nothing more to take out.
The area picture needs lengths that can be drawn. It can show 3(x + 4) = 3x + 12 for any width x, but it can show 5(x - 2) = 5x - 10 only when x is at least 2, so that there is a strip to cut off. That is a limit of the picture, not of the equation: the distributive property holds for every number, and later lessons on negative numbers use it when x is less than 2 too.
Where this goes next
The same three moves carry on with negative numbers. Distributing with variables puts a negative number outside the brackets and subtraction inside; combining like terms treats subtraction as adding the opposite; and distribute, then combine chains the two moves on longer expressions. Factoring out the greatest common factor takes the factor in Part 2 further, so that a letter can be part of it, as in 6x^2 + 9x = 3x(2x + 3).
Choosing which equivalent form to write is a skill of its own. Each form says something different about the same quantity, which is what reading the parts of an expression and rewriting an expression to show its meaning are about. And when you solve equations, many steps are rewrites of this kind: a side is replaced by an equivalent, simpler expression, which cannot change which numbers make the equation true.
Exercises
Part 1: distributing
Write each expression without brackets.
- 4(x + 9)
- 7(2a + 5)
- \displaystyle \frac{1}{2}(6k + 14)
- 8(y - 3)
- A rectangle is 6 units tall. A line splits it into two pieces, one p units wide and one 2 units wide. Write the area of the whole rectangle as a product and as a sum.
- Thor buys 5 cups of bubble tea. Each costs t dollars for the tea plus 0.8 dollars for jelly. Write the total cost as a product and as a sum without brackets.
Part 2: factoring out
Factor each expression fully, using the greatest common factor.
- 10x + 35
- 18b + 27
- 16 + 40n
- 11z + 11
- 12p + 30q + 6
- Find the missing number: \square(3h + 8) = 21h + 56.
- Write 56 + 35 as a product, using the greatest common factor of 56 and 35.
- A garden center has 48 pots of tomatoes and 36 pots of peppers to set out in identical rows, each with the same mix and nothing left over. What is the greatest number of rows? Write 48 + 36 as a product that shows the rows, and say what is in each row.
- Is 2(10x + 15) equivalent to 20x + 30? Is it fully factored? If not, factor 20x + 30 fully.
Part 3: like terms and testing
Write each expression with as few terms as possible.
- n + n + n + n + n
- 6c + 2c + 9
- 3d + 4 + d + 10
- 2x^2 + 3x + x^2 + 5x
- On a bird count, Uri’s tally sheet reads r + 2 + r + 2 + r + 2, for r robins and 2 finches at each of 3 stops. Combine like terms, then write the result as a product.
Decide whether each pair is equivalent, and prove your answer.
- 4(y + 3) and 4y + 3
- 5(x + 2) + 3x and 8x + 10
- x^2 + 2 and 3x. (Begin by checking x = 1 and x = 2.)
Reasoning
- Explain why trying x = 1, x = 2 and x = 3 and getting the same value from two expressions each time does not prove they are equivalent, but one value where they differ does prove they are not.
- Use the distributive property to explain why 4b + 3b = 7b for every b. Then explain why 4b + 3 cannot be combined into one term.
Answers
- 4x + 36. The 4 multiplies both terms: 4 \times x and 4 \times 9 = 36.
- 14a + 35. 7 \times 2a = (7 \times 2)a = 14a and 7 \times 5 = 35.
- 3k + 7. Half of 6k is 3k and half of 14 is 7.
- 8y - 24. 8 \times y less 8 \times 3 = 24.
- 6(p + 2) = 6p + 12. The whole rectangle is 6 by p + 2; the pieces have areas 6p and 12.
- 5(t + 0.8) = 5t + 4. One cup costs t + 0.8; distributing, 5 \times 0.8 = 4 dollars of jelly in all.
- 5(2x + 7). The greatest common factor of 10 and 35 is 5.
- 9(2b + 3). The greatest common factor of 18 and 27 is 9.
- 8(2 + 5n). The greatest common factor of 16 and 40 is 8. (8(5n + 2) is the same.)
- 11(z + 1). 11 = 11 \times 1, so a 1 is left in its place.
- 6(2p + 5q + 1). The greatest common factor of 12, 30 and 6 is 6.
- 7. The missing number times 3 is 21, and times 8 is 56; both give 7, and 7(3h + 8) = 21h + 56.
- 7(8 + 5). The greatest common factor of 56 and 35 is 7; both sides are 91.
- 12 rows, 48 + 36 = 12(4 + 3), each row with 4 tomatoes and 3 peppers. The greatest common factor of 48 and 36 is 12.
- Yes, it is equivalent: 2 \times 10x + 2 \times 15 = 20x + 30. It is not fully factored, because 10x and 15 still share a 5. Fully factored: 10(2x + 3).
- 5n. Five n’s added together is five times n.
- 8c + 9. 6c + 2c = (6 + 2)c = 8c; the 9 is not a like term.
- 4d + 14. 3d + d = 4d, since a lone d has coefficient 1, and 4 + 10 = 14.
- 3x^2 + 8x. 2x^2 + x^2 = 3x^2 and 3x + 5x = 8x; x^2 and x are not like terms.
- 3r + 6 = 3(r + 2). Three r’s and three 2’s; the factored form counts one stop, r + 2 birds, three times.
- Not equivalent. At y = 0, 4(0 + 3) = 12 but 4 \times 0 + 3 = 3. (Distributing gives 4y + 12.)
- Equivalent. 5(x + 2) + 3x = 5x + 10 + 3x = 8x + 10, using only the distributive property and combining like terms.
- Not equivalent. At x = 1 both are 3 and at x = 2 both are 6, but at x = 0, x^2 + 2 = 2 while 3x = 0.
- Equivalent means the two expressions give the same value for every number. Three values are only three of endless numbers, and two different expressions can agree at some values and not at others, as x^2 + 2 and 3x agree at 1 and 2 but not at 0. One value where they differ is a number for which “the same for every number” is false, so that one value is enough.
- The letter b is a factor of both terms, so the distributive property, read from right to left, gives 4b + 3b = (4 + 3)b = 7b. The property holds for every number, so this holds for every b: four b’s and three more b’s are seven b’s. In 4b + 3 the 3 has no factor b, so there is no common b to take out, and 4b counts b’s while 3 counts ones. It is not 7b: at b = 0, 4b + 3 = 3 but 7b = 0. It is not the plain number 7 either: at b = 0 it is 3, not 7.