Combining like terms
An expression mentions the same kind of thing several times over. How do you count those up into fewer terms without changing what the expression is worth?
Where you are 12 in Expressions
When you reach for it
Add up the three sides of a triangle whose sides are 3t + 2, 5t - 4 and 2t + 7, and you get
That is correct, but it is long: some t’s, some plain numbers, more t’s, more plain numbers, in no particular order. The same perimeter can be written as 10t + 5, two terms instead of six.
Rewriting an expression with as few terms as possible, by counting together the terms that are the same kind of thing, is combining like terms. You reach for it whenever an expression has been built by adding pieces together: a perimeter, a total of two kinds of item, the result of opening a bracket, either side of an equation. A shorter expression is quicker to evaluate, easier to compare with another one, and it is the form that solving an equation needs. The words term, coefficient and constant come from Terms, coefficients and constants.
This lesson runs over two days. Part 1 says which terms are alike, why they combine, and how to handle fraction and decimal coefficients. Part 2 brings in negative coefficients, several letters, and perimeters and totals.
The procedure
Part 1: like terms, and why they combine
The variable part of a term is everything in it except the number in front: in 7x^2 it is x^2, in \displaystyle \frac{1}{2}mn it is mn. Two terms are like terms when their variable parts are the same, letter for letter, with each letter raised to the same power. Only the coefficients may differ. All constants are like terms with one another, since none has a variable part.
| Pair | Like terms? | Why |
|---|---|---|
| 5x and \displaystyle \frac{1}{2}x | yes | both have variable part x |
| 5x and 5x^2 | no | x and x^2 are different powers |
| 3x and 3y | no | different letters |
| 4mn and 7nm | yes | nm = mn, by the commutative property of multiplication |
| a^2b and ab^2 | no | a \cdot a \cdot b and a \cdot b \cdot b are different products |
| 8 and 0.3 | yes | both are constants |
Why like terms can be combined. The distributive property says that for any numbers a, b and x,
Read it from right to left. It says that two terms with the same variable part can be rewritten as one term, whose coefficient is the sum of the two coefficients:
In a picture, x is a length you have not been told. Four copies of it and three more copies of it are seven copies of it, whatever the length is.
The same step works for any shared variable part: 2x^2 + 6x^2 = 8x^2 and 4mn + 7mn = 11mn. It does not work for unlike terms, because there is no shared variable part to pull out. The expression 4x + 3 is not 7x: at x = 10 the first is 4(10) + 3 = 43 and the second is 70. And 4x + 3x^2 stays as it is, two terms, because x and x^2 are different quantities.
Gathering like terms from anywhere. Like terms are rarely written next to each other. The commutative property of addition lets terms change places, and the associative property lets them be grouped in any way, so the like terms in an expression can always be brought together and then combined. An expression is simplified, as far as this lesson goes, when no two of its terms are like terms. By convention the answer is written with higher powers first and the constant last, as in 2c^2 + 4c + 9; that order is a habit that makes answers easy to compare, not a rule of algebra.
Hidden coefficients, fractions and decimals. A letter on its own has coefficient 1, so w = 1w and 7w + w = 8w. A letter divided by a number has a fraction for its coefficient, because dividing by 3 is multiplying by \displaystyle \frac{1}{3}, so \displaystyle \frac{y}{3} = \frac{1}{3}y. Fraction and decimal coefficients are numbers like any others, so the distributive property applies to them unchanged; adding them is ordinary fraction and decimal addition, with a common denominator where one is needed:
Checking. Combining like terms never changes the value of an expression, so the old and new expressions must give the same number for any value of the letter. Putting in one value, such as x = 2, catches almost every slip.
Part 2: negative coefficients and several letters
Subtraction is adding the opposite. Taking away a number gives the same result as adding its opposite, as shown in Subtracting integers. So every subtraction in an expression can be read as an addition of a negative term:
Read this way, the minus sign in front of a term belongs to that term. The terms of 6a - 2b - 4a are 6a, -2b and -4a. When the commutative property moves a term, its sign travels with it:
The coefficient 6 + (-4) = 2 is an ordinary sum of integers, worked out by the rules in Adding integers; with fractions or decimals it is a sum of rational numbers. A term written first with no sign in front of it has a plus sign understood.
Three consequences are worth seeing once.
- A term -x has coefficient -1, because -x = -1 \cdot x (the sign rules for products give -1 \cdot x as the opposite of x). So 5x - x = 5x + (-1x) = 4x.
- Like terms can cancel completely. 3k - 3k = \bigl(3 + (-3)\bigr)k = 0k = 0, and a term worth 0 is simply left out.
- Subtracting a negative term adds it: 5m - (-3m) = 5m + 3m = 8m.
Several letters. With more than one letter, sort the terms by variable part and combine each group on its own. Nothing ever crosses from one group to another: 4p + 3q is already as simple as it can be. Before sorting, write each variable part the same way, with letters in alphabetical order, so that 5nm is seen to be 5mn.
Worked examples
Examples 1 to 3 go with Part 1. Examples 4 to 7 go with Part 2.
Gathering like terms
Simplify (a) 7w + 4 + 5w + 9 \qquad (b) 3c + 2c^2 + c.
Solution(a) The terms are 7w, 4, 5w and 9. Two have variable part w and two are constants. The commutative property of addition lets 4 and 5w swap places, and the associative property lets the like terms be grouped:
By the distributive property, 7w + 5w = (7 + 5)w = 12w, and 4 + 9 = 13. So the expression is 12w + 13.
Check at w = 1: the original gives 7 + 4 + 5 + 9 = 25 and the answer gives 12 + 13 = 25.
(b) The term c has the hidden coefficient 1, so 3c and c are like terms: 3c + c = (3 + 1)c = 4c. The term 2c^2 has variable part c^2, which nothing else shares, so it stays as it is. Writing the higher power first, the result is 2c^2 + 4c.
Fraction and decimal coefficients
Simplify (a) \displaystyle \frac{3}{4}y + \frac{y}{6} \qquad (b) 1.6h + 2.75 + 0.9h + 0.5.
Solution(a) Dividing by 6 is multiplying by \displaystyle \frac{1}{6}, so \displaystyle \frac{y}{6} = \frac{1}{6}y, and the two terms are like terms. Their coefficients add over the common denominator 12:
(b) Gather the h terms and the constants, then add each group as decimals:
Check at h = 2: the original gives 3.2 + 2.75 + 1.8 + 0.5 = 8.25, and the answer gives 5 + 3.25 = 8.25.
Finding a missing term
Find the term that makes each statement true for every value of the letter.
(a) 6g + \square + 2 = 10g + 2 \qquad (b) \displaystyle \frac{2}{5}z + \square = z \qquad (c) \square + 2.5k^2 + k = 4k^2 + k
SolutionRun the combining backwards. The right side shows what each group of like terms must add up to, and the missing term supplies the difference.
(a) The constants already match, 2 on each side. The g terms must add to 10g, and 6g is there, so the missing term is a g term with coefficient 10 - 6 = 4. It is 4g: 6g + 4g + 2 = 10g + 2.
(b) The right side is z = 1z. The coefficients must add to 1, and \displaystyle 1 - \frac{2}{5} = \frac{3}{5}, so the missing term is \displaystyle \frac{3}{5}z.
(c) The k terms match. The k^2 terms must add to 4k^2, and 2.5k^2 is there, so the missing term is 1.5k^2. A plain 1.5 or 1.5k would not do, because neither is like k^2.
Subtracting like terms
Simplify (a) 8b - 3b \qquad (b) -5v + 2v \qquad (c) 4j - 9 - 7j + 2.
Solution(a) Subtracting is adding the opposite, so 8b - 3b = 8b + (-3b) = \bigl(8 + (-3)\bigr)b = 5b.
(b) The two coefficients are -5 and 2. Their sum is -3, because the numbers have different signs and 5 is the larger size, so -5v + 2v = -3v.
(c) Read every term with its sign: 4j, -9, -7j and +2. Move them so like terms sit together, each one taking its sign with it:
Check at j = 1: the original gives 4 - 9 - 7 + 2 = -10, and the answer gives -3 - 7 = -10.
Terms that cancel, and subtracting a negative
Simplify (a) 6e - e - 5e + 3 \qquad (b) \displaystyle \frac{1}{2}u - \frac{5}{6}u \qquad (c) 2.4s - (-1.6s) - 0.5.
Solution(a) The e terms are 6e, -e = -1e and -5e. Their coefficients add to 6 + (-1) + (-5) = 0, so together they are 0e = 0. Only the constant is left: the expression is 3, whatever e is.
Check at e = 10: 60 - 10 - 50 + 3 = 3.
(b) Over the common denominator 6, the coefficients are \displaystyle \frac{3}{6} and \displaystyle -\frac{5}{6}, and
so \displaystyle \frac{1}{2}u - \frac{5}{6}u = -\frac{1}{3}u. The result is negative because more is taken away than was there.
(c) Subtracting -1.6s is adding its opposite, 1.6s. So the expression is 2.4s + 1.6s - 0.5 = 4s - 0.5.
Several letters at once
Simplify (a) 9m - 4n + 2mn - 3m + n - 5nm \qquad (b) a^2b + 3ab^2 - 4ba^2.
Solution(a) First write -5nm as -5mn, since nm = mn. Then sort the terms by variable part, each with its sign:
- variable part m: 9m and -3m, which add to 6m;
- variable part n: -4n and n = 1n, which add to -3n;
- variable part mn: 2mn and -5mn, which add to -3mn.
So the expression is 6m - 3n - 3mn.
Check at m = 1, n = 2: the original gives 9 - 8 + 4 - 3 + 2 - 10 = -6, and the answer gives 6 - 6 - 6 = -6.
(b) The last term is -4ba^2 = -4a^2b, so it is like the first term, and a^2b + (-4a^2b) = -3a^2b. The term 3ab^2 has variable part a \cdot b \cdot b, which is not a \cdot a \cdot b, so it stays on its own. The result is -3a^2b + 3ab^2.
A perimeter and a total
(a) The sides of a triangle are 3t + 2, 5t - 4 and 2t + 7 centimeters. Write its perimeter as simply as possible.
(b) Verity buys 5 bags of marbles and 2 tubes of beads. Sebastiano buys 3 bags of marbles and 4 tubes of beads, then takes 1 bag of marbles back to the shop. Each bag holds m marbles and each tube holds b beads. Write the number of marbles and beads they have between them as simply as possible, and find it when m = 25 and b = 40.
Solution(a) The perimeter is the sum of the three sides. Nothing multiplies the brackets, so they can be dropped:
The perimeter is 10t + 5 centimeters.
Check at t = 3: the sides are 11, 11 and 13, which add to 35, and 10(3) + 5 = 35.
(b) Count each kind of item as a term: bags of marbles give terms in m, tubes of beads give terms in b, and the bag taken back is a term -m.
The coefficients now have a meaning: they keep 7 bags of marbles and 6 tubes of beads in all. When m = 25 and b = 40 that is 7(25) + 6(40) = 175 + 240 = 415 items. The answer cannot shrink further to 13m or 13b, because a bag and a tube need not hold the same number of things.
Boundaries
- Only like terms combine. The variable parts must match exactly, each letter to the same power. x and x^2 never combine, and neither do xy and x.
- A bracket is one term until it is opened. In 4(x + 1) + 3x the x inside the bracket is not yet a term of the whole expression, so it cannot be combined with 3x as it stands. The bracket has to be multiplied out first, which is distributing, then combining. A bracket with nothing multiplying it and only a plus sign in front, as in Example 7, can simply be dropped.
- Agreeing at one value is not being equal. Substituting a value checks a simplification, but it cannot prove one. x^2 + x and 2x^2 give the same value at x = 0 and at x = 1, yet at x = 2 they give 6 and 8. Two expressions are equivalent only when they agree for every value, which is what the properties used in combining guarantee.
- A length must come out positive. In Example 7 the simplification 10t + 5 holds for every t, but a triangle with those sides exists only for some values. Every side must be positive and, by the Triangle inequality, any two sides must add to more than the third. The pair 3t + 2 and 5t - 4 must beat 2t + 7, so 8t - 2 > 2t + 7, which holds exactly when \displaystyle t > \frac{3}{2}. The other two pairs, and the three sides being positive, all hold once \displaystyle t > \frac{3}{2}, so that is the whole condition.
Where this goes next
Combining is usually the second half of a rewrite. Opening brackets with the distributive property produces terms that then need combining, which is Distribute, then combine, and together the two moves are how equivalent expressions are made and recognized. Reading the parts of an expression as quantities starts from the simplified form, as the 7m + 6b of Example 7 did.
Solving an equation such as 3x + 2x - 4 = 11 starts by combining each side, which is Multi-step equations. Adding and subtracting polynomials, in Adding and subtracting polynomials, is combining like terms and nothing more. The idea also outlives letters: \displaystyle 2\sqrt{3} + 5\sqrt{3} = 7\sqrt{3} for the same reason 2x + 5x = 7x, which is part of working with radicals.
Exercises
Part 1: like terms
- For each pair, say whether they are like terms: (a) 4g and 4g^2 (b) \displaystyle \frac{1}{2}t and 9t (c) 3ab and 8ba (d) 6 and 0.5 (e) 2x^2y and 2xy^2.
Part 1: combining
Simplify.
- 6k + 9k
- 4 + 3z + 8 + z
- 2f^2 + 5f + 3f^2 + 2f
- 10h + 1 + h^2 + h
Part 1: fractions and decimals
Simplify.
- \displaystyle \frac{2}{5}a + \frac{1}{10}a
- \displaystyle \frac{c}{2} + \frac{2c}{5}
- 0.75d + 1.5 + 2.25d + 0.3
- Find the term that makes 5y + \square + 3 = 12y + 3 true for every y.
Part 2: negative coefficients
Simplify.
- 11q - 6q
- 3 - 7v + 5 + 2v
- -2x^2 + 6x - x^2 - 6x
- 9 - (-2b) - 3b
- \displaystyle \frac{3}{8}m - \frac{3}{4}m + 1
- -1.4k + 0.6 - 2.1k - 1.6
Part 2: several letters
Simplify.
- 4a - 3b + 2a + 5b
- 7gh - 2g + 3hg - 4h + g
- x^2y + 2xy^2 - 5yx^2 + xy
Part 2: perimeters and totals
- A pentagon has sides a, a + 6, 2a, 2a - 1 and 4a + 3 meters. Write its perimeter as simply as possible.
- Rhiannon buys 6 boxes of red tiles and 4 boxes of green tiles. Quillon buys 3 boxes of red tiles and 2 boxes of green tiles, and Rhiannon then returns 1 box of green tiles. A box of red tiles holds r tiles and a box of green tiles holds g tiles. Write the number of tiles they have between them as simply as possible, and find it when r = 20 and g = 12.
- A rectangle is 4w - 3 centimeters long and w + 2 centimeters wide. Add its four sides to write its perimeter as simply as possible, and check your answer when w = 5.
Reasoning
- Explain why 5y - 2 cannot be simplified to 3y, and give a value of y that shows the two are different.
- Show that 8p - 3q - 5p equals 3p - 3q. Name the property that lets -5p move next to 8p, and say what happens to its minus sign when it moves.
- The expressions x^2 + 2 and 3x give the same value at x = 1 and at x = 2. Are they equivalent? Explain.
- Explain why ax - bx = (a - b)x for every pair of numbers a and b, using subtraction as adding the opposite and the distributive property.
Answers
- (a) No: g and g^2 are different powers. (b) Yes: both have variable part t. (c) Yes: ba = ab. (d) Yes: both are constants. (e) No: x^2y is x \cdot x \cdot y and xy^2 is x \cdot y \cdot y.
- 15k. 6k + 9k = (6 + 9)k.
- 4z + 12. The z terms are 3z + 1z = 4z and the constants are 4 + 8 = 12.
- 5f^2 + 7f. 2f^2 + 3f^2 = 5f^2 and 5f + 2f = 7f.
- h^2 + 11h + 1. 10h + h = 11h; h^2 and 1 have nothing to combine with.
- \displaystyle \frac{1}{2}a. \displaystyle \frac{2}{5} = \frac{4}{10}, and \displaystyle \frac{4}{10} + \frac{1}{10} = \frac{5}{10} = \frac{1}{2}.
- \displaystyle \frac{9}{10}c. \displaystyle \frac{c}{2} = \frac{5}{10}c and \displaystyle \frac{2c}{5} = \frac{4}{10}c, which add to \displaystyle \frac{9}{10}c.
- 3d + 1.8. 0.75 + 2.25 = 3 and 1.5 + 0.3 = 1.8.
- 7y. The constants match, and the y coefficients must add to 12, so 12 - 5 = 7.
- 5q. 11q + (-6q) = \bigl(11 + (-6)\bigr)q.
- -5v + 8. -7v + 2v = -5v and 3 + 5 = 8.
- -3x^2. -2x^2 + (-1x^2) = -3x^2, and 6x + (-6x) = 0, so the x terms cancel.
- -b + 9. Subtracting -2b adds 2b: 9 + 2b - 3b = 9 + (-1b).
- \displaystyle -\frac{3}{8}m + 1. \displaystyle \frac{3}{4} = \frac{6}{8}, and \displaystyle \frac{3}{8} + \left(-\frac{6}{8}\right) = -\frac{3}{8}.
- -3.5k - 1. -1.4 + (-2.1) = -3.5 and 0.6 + (-1.6) = -1.
- 6a + 2b. 4a + 2a = 6a and -3b + 5b = 2b.
- 10gh - g - 4h. hg = gh, so 7gh + 3gh = 10gh; -2g + 1g = -g; -4h has no partner.
- -4x^2y + 2xy^2 + xy. yx^2 = x^2y, so x^2y - 5x^2y = -4x^2y; 2xy^2 and xy are unlike everything else.
- 10a + 8 meters. a + a + 2a + 2a + 4a = 10a and 6 + (-1) + 3 = 8.
- 9r + 5g tiles, which is 240 when r = 20 and g = 12. Red: 6 + 3 = 9 boxes; green: 4 + 2 - 1 = 5 boxes. 9(20) + 5(12) = 180 + 60 = 240.
- 10w - 2 centimeters. The four sides are 4w - 3, w + 2, 4w - 3 and w + 2. The w terms add to 10w and the constants to -3 + 2 - 3 + 2 = -2. At w = 5 the sides are 17 and 7, and 17 + 7 + 17 + 7 = 48, while 10(5) - 2 = 48.
- 5y and -2 are not like terms: -2 has no variable part, so there is no shared variable part to pull out with the distributive property. At y = 10, 5y - 2 = 48 but 3y = 30.
- Read the terms with their signs: 8p, -3q, -5p. The commutative property of addition lets -5p move next to 8p, and it takes its minus sign with it, because the sign is part of the term: 8p + (-5p) + (-3q) = 3p + (-3q) = 3p - 3q.
- No. Equivalent expressions agree for every value of x, and these do not: at x = 3, x^2 + 2 = 11 but 3x = 9. Agreeing at two values is not enough.
- Subtracting is adding the opposite, so ax - bx = ax + (-b)x. The distributive property, read right to left, gives ax + (-b)x = \bigl(a + (-b)\bigr)x, and a + (-b) = a - b. Every step holds for any numbers a, b and x, so ax - bx = (a - b)x always.