Chapter 6 ยท Equations and inequalities

Combining Like Terms and Solving Word Problems

Combine like terms to turn a longer equation into a two-step equation, then build and solve equations from stories.

20 min read

You can already solve a two-step equation such as 5x+4=395x + 4 = 39. In this lesson the variable appears more than once, as in 3x+2x+4=393x + 2x + 4 = 39, and later the equation is hidden inside a story. The key new move is to tidy the equation first, then solve it the way you know.

Three words will help.

  • The terms of an expression are the parts that are added together. Subtracting counts as adding the opposite, so the terms of 4yโˆ’9โˆ’6y4y - 9 - 6y are 4y4y, โˆ’9-9 and โˆ’6y-6y. Each term keeps the sign in front of it.
  • A coefficient is the number that multiplies the variable in a term. In 4y4y it is 44, and in โˆ’6y-6y it is โˆ’6-6. A variable standing alone has the coefficient 11, because n=1โ‹…nn = 1 \cdot n.
  • A constant is a term with no variable, such as โˆ’9-9.

Like terms are terms whose variable part is exactly the same. In this lesson that means the same letter, as in 3x3x and 2x2x, or no letter at all, as in the constants โˆ’9-9 and 22. But 3x3x and 44 are not like terms: 3x3x counts groups of xx, and 44 counts single units.

Why can you combine like terms? Think of xx as the size of one group. Then 3x3x is 3 groups and 2x2x is 2 groups, so together they make 5 groups of that size: 3x+2x=(3+2)x=5x3x + 2x = (3 + 2)x = 5x. This is the distributive property, and it holds whatever number xx stands for.

Two expressions are equivalent only if they are equal for every value of the variable. So 3x+2x3x + 2x and 5x5x are equivalent. The table below can test only a few values of xx. The distributive property is what guarantees the match for every value.

Testing 3x + 2x and 5x with the same values of x
Value of xValue of 3x + 2xValue of 5x
โˆ’2-2
3(โˆ’2)+2(โˆ’2)=โˆ’6+(โˆ’4)=โˆ’103(-2) + 2(-2) = -6 + (-4) = -10
5(โˆ’2)=โˆ’105(-2) = -10
00
3(0)+2(0)=0+0=03(0) + 2(0) = 0 + 0 = 0
5(0)=05(0) = 0
1.51.5
3(1.5)+2(1.5)=4.5+3=7.53(1.5) + 2(1.5) = 4.5 + 3 = 7.5
5(1.5)=7.55(1.5) = 7.5
44
3(4)+2(4)=12+8=203(4) + 2(4) = 12 + 8 = 20
5(4)=205(4) = 20
77
3(7)+2(7)=21+14=353(7) + 2(7) = 21 + 14 = 35
5(7)=355(7) = 35

Key concept

Combine like terms, then solve in two steps

Like terms count groups of the same thing, so you can add or subtract their coefficients. The new expression is equivalent to the old one, which means equal for every value of the variable. That is why the equation keeps the same solutions, the values that make it true.

After combining, each equation in this lesson has one variable term, whose coefficient is not zero, and one constant on the same side. Solve it by doing the same operation to both sides: add or subtract to leave the variable term alone, then divide by its coefficient. Dividing is allowed only because that coefficient is not zero.

A solution must make the original equation true, so check your value there.

ax+bx=(a+b)xax + bx = (a + b)x
Combining like terms: each term keeps the sign in front of it
ExpressionLike termsCombine themEquivalent expression
6a+3aโˆ’26a + 3a - 2
6a6a and 3a3a
(6+3)a=9a(6 + 3)a = 9a
9aโˆ’29a - 2
5bโˆ’8b+15b - 8b + 1
5b5b and โˆ’8b-8b
(5โˆ’8)b=โˆ’3b(5 - 8)b = -3b
โˆ’3b+1-3b + 1
n+2.5nโˆ’3n + 2.5n - 3
nn and 2.5n2.5n, where n=1โ‹…nn = 1 \cdot n
(1+2.5)n=3.5n(1 + 2.5)n = 3.5n
3.5nโˆ’33.5n - 3
7+6kโˆ’27 + 6k - 2
77 and โˆ’2-2
7โˆ’2=57 - 2 = 5
6k+56k + 5
10โˆ’4mโˆ’2m10 - 4m - 2m
โˆ’4m-4m and โˆ’2m-2m
(โˆ’4โˆ’2)m=โˆ’6m(-4 - 2)m = -6m
10โˆ’6m10 - 6m

Worked example

Combine, then solve

Solve 3x+2x+4=393x + 2x + 4 = 39.

  1. Find the like terms. 3x and 2x both have the variable x, so they are like terms. The constant 4 has no variable, so it stays separate.

    3x+2x+4=393x + 2x + 4 = 39
  2. Combine the like terms by adding their coefficients: 3 groups of x and 2 groups of x make 5 groups of x. The new left side equals the old one for every value of x, so the equation keeps the same solutions.

    (3+2)x+4=39so5x+4=39(3 + 2)x + 4 = 39 \quad \text{so} \quad 5x + 4 = 39
  3. Subtract 4 from both sides to undo adding 4. The two sides were equal, so doing the same operation to both keeps them equal.

    5x+4โˆ’4=39โˆ’4so5x=355x + 4 - 4 = 39 - 4 \quad \text{so} \quad 5x = 35
  4. Divide both sides by 5 to undo multiplying by 5. This is allowed because 5 is not zero, and doing it to both sides keeps them equal.

    5x5=355sox=7\frac{5x}{5} = \frac{35}{5} \quad \text{so} \quad x = 7
  5. Check in the original equation, not the simplified one, because a solution must make the original equation true. Replace x with 7.

    3(7)+2(7)+4=21+14+4=393(7) + 2(7) + 4 = 21 + 14 + 4 = 39

Answer

The solution is x=7x = 7, because it makes the original equation 3x+2x+4=393x + 2x + 4 = 39 true.

Now for stories. Each story in this lesson describes one amount in two ways: built from its parts, and as a number the story gives you. An equation says the two descriptions are equal. Four moves turn the sentences into symbols.

  1. Name the unknown. Choose a variable and write exactly what it stands for, with its unit, such as the cost of one ticket in dollars.
  2. Build the equation. Write a term for each part of the story. Add the parts that make the amount larger and subtract the parts that make it smaller. Set the result equal to the number the story gives.
  3. Solve. Combine like terms, then do the same operation to both sides. Divide only by a number that is not zero.
  4. Check. A solution must make the original equation true. Then test it in the story's own words, because a value can fit your equation and still be wrong if the equation was built incorrectly.

Worked example

A story with the same unknown twice

Maya walks dogs and is paid the same amount for each hour of work. One weekend she works 3.53.5 hours on Saturday and 2.52.5 hours on Sunday. A customer also gives her a tip of 77 dollars. She earns 7979 dollars in all. How much is Maya paid for one hour?

  1. Name the unknown with its unit. Let r stand for Maya's pay for one hour, in dollars per hour. Hours times dollars per hour gives dollars, so Saturday's pay is 3.5r dollars and Sunday's pay is 2.5r dollars. The tip is a constant, 7 dollars.

    Saturday:ย 3.5rSunday:ย 2.5rTip:ย 7\text{Saturday: } 3.5r \qquad \text{Sunday: } 2.5r \qquad \text{Tip: } 7
  2. Build the equation. Saturday's pay plus Sunday's pay plus the tip is everything Maya earned, and the story says that amount is 79 dollars. Every term is an amount in dollars, so the terms can be added.

    3.5r+2.5r+7=793.5r + 2.5r + 7 = 79
  3. Combine the like terms 3.5r and 2.5r. They are like terms because r is the same hourly pay on both days. Add the coefficients: 3.5 + 2.5 = 6. The tip, 7, is a constant, so it stays separate. The new left side equals the old one for every value of r.

    (3.5+2.5)r+7=79so6r+7=79(3.5 + 2.5)r + 7 = 79 \quad \text{so} \quad 6r + 7 = 79
  4. Subtract 7 from both sides so the variable term is alone. Doing the same operation to both sides keeps them equal.

    6r+7โˆ’7=79โˆ’7so6r=726r + 7 - 7 = 79 - 7 \quad \text{so} \quad 6r = 72
  5. Divide both sides by 6, which is allowed because 6 is not zero.

    6r6=726sor=12\frac{6r}{6} = \frac{72}{6} \quad \text{so} \quad r = 12
  6. Check in the original equation and in the story. Saturday's pay is 3.5 times 12, which is 42 dollars. Sunday's pay is 2.5 times 12, which is 30 dollars. With the 7-dollar tip, the total is 79 dollars, as the story says. A pay of 12 dollars per hour is positive, so it makes sense.

    3.5(12)+2.5(12)+7=42+30+7=793.5(12) + 2.5(12) + 7 = 42 + 30 + 7 = 79

Answer

Maya is paid 1212 dollars for one hour of dog walking.

Worked example

A story where the amount goes down

Kai has a gift card worth 5050 dollars. He uses it to buy 33 game passes on Friday and 22 more game passes on Saturday. Every pass costs the same amount. Then 1515 dollars are left on the card. How much does one game pass cost?

  1. Name the unknown with its unit: let c stand for the cost of one game pass, in dollars. Friday's passes cost 3c dollars and Saturday's passes cost 2c dollars. Spending makes the balance smaller, so subtract both amounts from the 50 dollars on the card. The story says 15 dollars are left.

    50โˆ’3cโˆ’2c=1550 - 3c - 2c = 15
  2. Rewrite each subtraction as adding the opposite. This shows that the like terms are โˆ’3c and โˆ’2c, each with its own sign. The left side keeps the same value for every value of c.

    50+(โˆ’3c)+(โˆ’2c)=1550 + (-3c) + (-2c) = 15
  3. Combine the like terms by adding their coefficients: โˆ’3 + (โˆ’2) = โˆ’5. Adding โˆ’5c is the same as subtracting 5c. The new left side equals the old one for every value of c.

    50+(โˆ’3+(โˆ’2))c=15so50โˆ’5c=1550 + (-3 + (-2))c = 15 \quad \text{so} \quad 50 - 5c = 15
  4. Subtract 50 from both sides so the variable term is alone. Doing the same operation to both sides keeps them equal. The negative sign stays with 5c.

    50โˆ’5cโˆ’50=15โˆ’50soโˆ’5c=โˆ’3550 - 5c - 50 = 15 - 50 \quad \text{so} \quad -5c = -35
  5. Divide both sides by โˆ’5, which is allowed because โˆ’5 is not zero. A negative number divided by a negative number is positive.

    โˆ’5cโˆ’5=โˆ’35โˆ’5soc=7\frac{-5c}{-5} = \frac{-35}{-5} \quad \text{so} \quad c = 7
  6. Check in the original equation and in the story. Three passes cost 21 dollars and two passes cost 14 dollars. Taking both amounts from 50 dollars leaves 15 dollars, as the story says. A cost of 7 dollars is positive, so it makes sense.

    50โˆ’3(7)โˆ’2(7)=50โˆ’21โˆ’14=1550 - 3(7) - 2(7) = 50 - 21 - 14 = 15

Answer

One game pass costs 77 dollars.

Takeaway. When a variable appears in more than one term, combine the like terms first. You may do this because the new expression is equivalent to the old one, which means equal for every value of the variable. Then solve the two-step equation by doing the same operation to both sides, dividing only by a number that is not zero. Check your value in the original equation.

A story works the same way once you name the unknown with its unit and build the equation from the story's parts. Check your answer in the story's words too.

Next, you will compare solving a problem with an equation to solving it with arithmetic alone, and you will look for errors in solutions of two-step equations.