7.5: Solving TwoStep Equations
Introduction
A Trip to Fenway Park
Marc is going to get his wish. After much ado, his Grandpa managed to pull a few strings and get tickets for the Red Sox game. Marc can hardly wait for Saturday’s game.
“I hope that it is as good as I think it will be,” Marc said.
“Well for $535.00 I hope we don’t get rained out.”
“Five hundred thirtyfive dollars!! Wow, that is a lot of money!” Marc exclaimed.
“Yes, but that does include a $15.00 service fee. I was able to secure four tickets and that is all that matters. Don’t worry Marc; having your dream come true is worth the money.”
“Thanks Grandpa,” Marc said, smiling.
Afterwards, Marc began to wonder what the price of each ticket really was. He knew that they got field box seats and that they had to be pretty expensive given the final bill. How could he figure out the ticket price?
This is where a two step equation will help Marc. This lesson will teach you all about writing and solving twostep equations. When finished, you will know how to figure out the ticket price for Marc.
What You Will Learn
In this lesson you will learn how to complete the following:
 Write twostep single variable equations from verbal models.
 Solve twostep single variable equations.
 Identify properties used to solve multistep equations.
 Model and solve realworld problems using twostep equations.
Teaching Time
I. Write TwoStep Single Variable Equations from Verbal Models
In the last two lessons you worked on solving onestep single variable equations. Here is an example:
Example
\begin{align*}3x=15\end{align*}
Now if you think about this equation, you only had to perform one operation to solve it. We used an inverse or opposite operation to solve it. Because this is a multiplication equation, we used division to solve for the unknown variable. We could say that we performed one step to solve this equation. The one step was to divide. We call an equation where one operation is needed to solve for the unknown variable a onestep equation.
What happens when there is more than one operation needed to solve an equation?
When this happens, we have twostep equation. A twostep equation has two operations in it. Let’s look at a few examples of twostep equations.
\begin{align*}3x+5 &= 20\\
\frac{x}{3}2 &= 5\end{align*}
If you look at both of these equations, you will see that there are two operations present in each. The first equation has multiplication and addition. The second equation has division and subtraction. You will see other variations of twostep equations too, but this gives you an idea.
Before we begin solving equations, let’s look at how we can write a twostep equation from a verbal model.
First, let’s think about some of the words that mean addition, subtraction, multiplication and division. Identifying these key words is going to assist us when writing equations from verbal models. Remember that a verbal model uses words.
Addition  Subtraction  Multiplication  Division 

Sum Added Altogether In all Plus and 
Difference Less than More than Take away subtract 
Product Times Groups of 
Quotient Split Up Divided 
Take a few minutes to write down these key words in your notebook.
Now let’s look at an example.
Example
Six times a number, plus five, is fortyone.
First, identify any key words that identify operations.
Six times a number, plus five, is fortyone.
Next, begin to translate the words into an equation.
Six = 6
Times \begin{align*}= "."\end{align*}
A number = x
Plus = +
Five = 5
Is means =
Fortyone is 41
Next, we put it altogether.
\begin{align*}6.x+5=41\end{align*}
Example
Four less than two times a number is equal to eight.
First, identify any key words that identify operations.
Four less than two times a number is eight.
Now we can translate each part.
Four becomes 4
Less than means subtraction
Two becomes 2
Times \begin{align*}= "."\end{align*}
A number = x
Is equal to means =
Eight = 8
Notice that the one tricky part is in the words “less than” because it is less than two times a number, the two times a number needs to come first in the equation. Then we can subtract.
Put it altogether.
\begin{align*}2.x4=8\end{align*}
Example
A number divided by two, plus six, is equal to fourteen.
First, look for key words that identify operations.
A number divided by two, plus six, is equal to fourteen.
Next, translate each word into an equation.
A number means variable (x)
Divided means \begin{align*}\div\end{align*}
By two means 2 is our divisor
Plus means addition
Six means 6
Is means =
Fourteen means 14
Put it all together.
\begin{align*}\frac{x}{2}+6=14\end{align*}
7K. Lesson Exercises
Write an equation for each.
 The product of five and a number, plus three, is twentythree.
 Six times a number, minus four, is thirtytwo.

A number, \begin{align*}y\end{align*}
y , divided by 3, plus seven, is ten.
Take a few minutes to check your work with a partner. Do your equations represent the verbal model?
II. Solve TwoStep Single Variable Equations
Now that you know how to write a twostep single variable equation it is time to learn how to solve them. When we solve the equation, we will be looking for the value of the variable in the equation.
Sometimes, you will be able to solve a twostep equation using mental math. Particularly when the numbers are small and are all positive integers, you might be able to solve them in your head. This is the first method that we are going to look at.
Example
\begin{align*}3x+3=9\end{align*}
Here the numbers are small. You can probably look at this one and say to yourself, “What number times three plus three is nine?” The logical answer is 2.
You can check your work by substituting 2 in for \begin{align*}x\end{align*}
\begin{align*}3(2)+3 &= 9\\
6+3 &= 9\\
9 &= 9\end{align*}
The answer checks out. The value of \begin{align*}x\end{align*}
That is a great question! This is where our second strategy comes in.
When you have a twostep equation that you can’t figure out in your head, you can use the strategy isolate the variable.
Example
\begin{align*}2+3n=11\end{align*}
Notice that there are two terms on the left side of the equation, 2 and \begin{align*}3n\end{align*}. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}3n\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 2 is added to \begin{align*}3n\end{align*}. So we can use the inverse of addition, which is subtraction, and subtract 2 from both sides of the equation. We need to subtract 2 from both sides of the equation because of the Subtraction Property of Equality. You probably remember that property states that in order to keep the values on both sides of the equation equal, whatever we subtract from one side of the equation must also be subtracted from the other side.
Let's see what happens when we subtract 2 from both sides of the equation.
\begin{align*}2 + 3n &= 11\\ 2  2+3n &= 112\\ 0+3n &= 9\\ 3n &= 9\end{align*}
Now, the term that includes a variable, \begin{align*}3n\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}n\end{align*} by itself. Since \begin{align*}3n\end{align*} means \begin{align*}3 \times n\end{align*}, we can use the inverse of multiplication––division. We can divide both sides of the equation by 3. We need to divide both sides of the equation by 3 because of the Division Property of Equality. That property states that if we divide one side of the equation by a number, we must divide the other side of the equation by the same number in order to keep the values on both sides of the equation equal.
Let's see what happens when we divide both sides of the equation by 3.
\begin{align*}3n &= 9\\ \frac{3n}{3} &= \frac{9}{3}\\ 1n &= 3\\ n &= 3\end{align*}
The value of \begin{align*}n\end{align*} is 3.
Example
\begin{align*}2x5=11\end{align*}
First, notice that there are two terms on the left side of the equal sign. To solve this equation, we want to isolate the variable and get \begin{align*}x\end{align*} alone. Let’s start with the term that is not connected to the \begin{align*}x\end{align*}, "5".
We use the Addition Property of Equality to cancel out the 5. We do this by using the inverse operation and adding 5 to both sides of the equation. Remember because of the properties of equality what we do to one side of the equals must match on the other side.
\begin{align*}2x5+5 &= 11+5\\ 2x+0 &= 16\\ 2x &= 16\end{align*}
Now we have a problem that is a multiplication problem. We can use the Division Property of Equality to solve for \begin{align*}x\end{align*} by dividing both sides of the equation by 2. Think about this and it makes perfect sense. Since we want to figure out what number times two is sixteen we can divide sixteen by two to figure it out.
\begin{align*}\frac{2x}{2} &= \frac{16}{2}\\ x &= 8\end{align*}
The value of \begin{align*}x\end{align*} is 8.
Example
\begin{align*}\frac{x}{5}8=17\end{align*}.
Notice that there are two terms on the left side of the equation, \begin{align*}\frac{x}{5}\end{align*} and 8. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}\frac{x}{5}\end{align*}, by itself on one side of the equal (=) sign.
In the equation, 8 is subtracted from \begin{align*}\frac{x}{5}\end{align*}. So, we can use the inverse of subtraction, which is addition, and add 8 to both sides of the equation. We need to add 8 to both sides of the equation in order to keep the values on both sides of the equation equal. Remember the Addition Property of Equality? Whatever we add to one side of the equation must also be added to the other side.
Let's see what happens when we add 8 to both sides of the equation.
\begin{align*}\frac{x}{5}8 &= 17\\ \frac{x}{5}8+8 &= 17+8\\ \frac{x}{5}+(8+8) &= 25\\ \frac{x}{5}+0 &= 25\\ \frac{x}{5} &= 25\end{align*}
Now, the term that includes a variable, \begin{align*}\frac{x}{5}\end{align*}, is by itself on one side of the equation.
We can now use inverse operations to get the \begin{align*}x\end{align*} by itself. Since \begin{align*}\frac{x}{5}\end{align*} means \begin{align*}x \div 5\end{align*}, we can use the inverse of division: multiplication. We can multiply both sides of the equation by 5. We need to multiply both sides of the equation by 5 because of the Multiplication Property of Equality.; multiply both sides of the equation by the same number in order to keep the values on both sides of the equation equal.
Let's see what happens when we multiply both sides of the equation by 5.
\begin{align*}\frac{x}{5} &= 25\\ \frac{x}{5} \times 5 &= 25 \times 5\\ \frac{x}{\cancel{5}} \times \frac{\cancel{5}}{1} &= 125\\ \frac{x}{1} &= 125\\ x &= 125\end{align*}
The value of \begin{align*}x\end{align*} is 125.
7L. Lesson Exercises
 \begin{align*}5x+7=32\end{align*}
 \begin{align*}3a+9=39\end{align*}
 \begin{align*}\frac{y}{4}8=4\end{align*}
Go over your answers with a partner. Correct any errors and then continue with the next section.
III. Using Properties
While solving twostep equations we have been utilizing the different properties of equality. These properties are valuable because they allow us to keep things in balance and with equations that is what it's all about. The quantities on each side of the equation must be kept equal.
Let’s review those properties and when you use them.
Subtraction Property of Equality is used when you have an equation with addition in it. It states that you can subtract the same quantity from both sides of an equation and it will still balance.
Addition Property of Equality is used when you have an equation with subtraction in it. It states that you can add the same quantity to both sides of an equation and it will still balance.
Division Property of Equality is used when you have an equation with multiplication in it. It states that you can divide the quantities on both sides of an equation by the same value and the equation will still balance.
Multiplication Property of Equality is used when you have an equation with division in it. It states that you can multiply the quantities on both sides of an equation by the same value and the equation will still balance.
As you continue to work on equations be sure to keep these properties in mind. They will be useful to help you remember which operations solve equations.
IV. Model and Solve RealWorld Problems Using TwoStep Equations
Now let’s apply what we have learned to solving realworld equations.
Example
A landscaper charges $35 for each landscaping job plus $20 for each hour worked. She charged $95 for one landscaping job.
a. Write an algebraic equation to represent \begin{align*}h\end{align*}, the number of hours that the landscaper worked on that $95 job.
b. How many hours did that job take?
Consider part \begin{align*}a\end{align*} first.
Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The landscaper earned $20 for each hour worked on that job, so you could multiply $20 by \begin{align*}h\end{align*}, the number of hours worked, to find how the landscaper charged for her worktime.
\begin{align*}& \underline{\$35} \ for \ each \ landscaping \ job \ \underline{plus} \ \underline{\$20 \ for \ each \ hour} \ worked \ldots \underline{charged} \ \underline{\$95} \ for \ one \ldots job.\\ & \ \downarrow \qquad \qquad \qquad \qquad \qquad \qquad \quad \downarrow \qquad \qquad \ \downarrow \qquad \qquad \qquad \qquad \qquad \quad \ \downarrow \qquad \downarrow\\ & \ 35 \qquad \qquad \qquad \qquad \qquad \quad \quad \ + \qquad \quad \ \ 20 h \qquad \qquad \qquad \qquad \qquad \ = \quad \ 95\end{align*}
So, this equation, \begin{align*}35+20h=95\end{align*}, represents \begin{align*}h\end{align*}, the number of hours the landscaper worked on the $95 job.
Next, consider part \begin{align*}b\end{align*}.
Solve the equation to find the number of hours the landscaper worked on that job.
First, subtract 35 from each side.
\begin{align*}35+20h &= 95\\ 3535+20h &= 9535\\ 0+20h &= 60\\ 20h &= 60\end{align*}
Next, divide both sides by 20.
\begin{align*}20h &= 60\\ \frac{20h}{20} &= \frac{60}{20}\\ 1h &= 3\\ h &= 3\end{align*}
The landscaper worked for 3 hours on the $95 job.
Now let’s apply what we have learned to solving the problem in the introduction.
Real Life Example Completed
A Trip to Fenway Park
Here is the original problem once again. Reread it and underline any important information.
Marc is going to get his wish. After much ado, his Grandpa managed to pull a few strings and get tickets for the Red Sox game. Marc can hardly wait for Saturday’s game.
“I hope that it is as good as I think it will be,” Marc said.
“Well for $535.00 I hope we don’t get rained out.”
“Five hundred thirtyfive dollars!! Wow, that is a lot of money!” Marc exclaimed.
“Yes, but that does include a $15.00 service fee. I was able to secure four tickets and that is all that matters. Don’t worry Marc; having your dream come true is worth the money.”
“Thanks Grandpa,” Marc said, smiling.
Afterwards, Marc began to wonder what the price of each ticket really was. He knew that they got field box seats and that they had to be pretty expensive given the final bill. How could he figure out the ticket price?
To figure out the ticket price we need to write and solve a twostep equation. Let’s look at what we know.
4 tickets at \begin{align*}x\end{align*} cost
+$15.00 service fee
= $535.00
Now let’s write the equation.
\begin{align*}4x+15=535\end{align*}
We start by subtracting 15 from both sides.
\begin{align*}4x+1515 &= 53515\\ 4x &= 520\\ \frac{4x}{4} &= \frac{520}{4}\\ x &= \$130.00\end{align*}
Each ticket costs $130.00.
Vocabulary
Here are the vocabulary words that are found in this lesson.
 Inverse Operation
 the opposite operation
 OneStep Equation
 an equation with one operation in it
 TwoStep Equation
 an equation with two operations in it
 Isolate the Variable
 a process to get the variable alone on one side of the equals.
 Subtraction Property of Equality
 is used when you have an equation with addition in it. It states that you can subtract the same quantity from both sides of an equation and it will still balance.
 Addition Property of Equality
 is used when you have an equation with subtraction in it. It states that you can add the same quantity to both sides of an equation and it will still balance.
 Division Property of Equality
 is used when you have an equation with multiplication in it. It states that you can divide the quantities on both sides of an equation by the same value and the equation will still balance.
 Multiplication Property of Equality
 is used when you have an equation with division in it. It states that you can multiply the quantities on both sides of an equation by the same value and the equation will still balance.
Technology Integration
Khan Academy Solving for a Variable
James Sousa, Solving Two Step Equations, The Basics
Other Videos:
 http://www.mathplayground.com/mv_solving_two_step_equations.html – This is a Brightstorm video on solving twostep equations.
Time to Practice
Directions: Write the following twostep equations from verbal models.
1. Two times a number, plus seven, is nineteen.
2. Three times a number, plus five, is twenty.
3. Six times a number, added to ten, is fortysix.
4. Seven less than two times a number is twentyone.
5. Eight less than three times a number is sixteen.
6. A number divided by two, plus seven, is ten.
7. A number divided by three and then combined with six is eleven.
8. Two less than a number divided by four is ten.
Directions: Solve each twostep equation for the unknown variable.
9. \begin{align*}3x+2=14\end{align*}
10. \begin{align*}6y+5=29\end{align*}
11. \begin{align*}7x+3=24\end{align*}
12. \begin{align*}5x+7=42\end{align*}
13. \begin{align*}6y+1=43\end{align*}
14. \begin{align*}9a+7=88\end{align*}
15. \begin{align*}11b+12=56\end{align*}
16. \begin{align*}12x3=21\end{align*}
17. \begin{align*}4y5=19\end{align*}
18. \begin{align*}3a9=21\end{align*}
19. \begin{align*}5b8=37\end{align*}
20. \begin{align*}7x10=39\end{align*}
Directions: Write an expression and/or solve each problem.
21. Augusta sells tshirts at the school store. On Tuesday, Augusta sold 7 less than twice the number of tshirts she sold on Monday. She sold 3 tshirts on Tuesday. Write an algebraic equation to represent \begin{align*}m\end{align*}, the number of tshirts August sold on Monday.
22. There are 19 green marbles in a box. The number of green marbles in the box is 6 more than half the number of red marbles in the box. Write an algebraic equation to represent \begin{align*}r\end{align*}, the number of red marbles in the box.
23. For each babysitting job he takes on, Manuel charges $2 for bus fare plus an additional $6 for each hour he works at the job. On Saturday, Manuel earned $26 for one babysitting job.
a. Write an equation to represent \begin{align*}h\end{align*}, the total number of hours that Manuel babysat during his $26 job.
b. Determine the total number of hours that Manuel babysat during that job.
24. Sophia has 29 trading cards in her collection. The number of cards in Sophia's collection is equal to 1 less than onethird of the number of trading cards in her brother Eli's collection.
a. Write an equation to represent \begin{align*}e\end{align*}, the number of trading cards in Eli's collection.
b. Determine the number of trading cards in Eli's collection.
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