# 7.12: Two-Step Equations and Properties of Equality

**At Grade**Created by: CK-12

**Practice**Two-Step Equations and Properties of Equality

Have you ever been to a Red Sox game?

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 thirty-five 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 Concept will teach you all about writing and solving two-step equations. When finished, you will know how to figure out the ticket price with Marc.**

### Guidance

Now that you know how to write a two-step 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 two-step equation using mental math. 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.**

\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*} x is 2.**

**That is a great question! This is where our second strategy comes in.**

**When you have a two-step equation that you can’t figure out in your head, you must use the strategy** *isolate the variable.*

\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*} 3n. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}3n\end{align*}3n, by itself on one side of the equal (=) sign.**

**In the equation, 2 is** *added***to \begin{align*}3n\end{align*} 3n. 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

**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.**

*Subtraction Property of Equality.*Let's see what happens when we subtract 2 from both sides of the equation.

\begin{align*}2 + 3n &= 11\\
2 - 2+3n &= 11-2\\
0+3n &= 9\\
3n &= 9\end{align*}

**Now, the term that includes a variable, \begin{align*}3n\end{align*} 3n, is by itself on one side of the equation.**

**We can now use inverse operations to get the \begin{align*}n\end{align*} n by itself. Since \begin{align*}3n\end{align*}3n means \begin{align*}3 \times n\end{align*}3×n, 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

**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.**

*Division Property of Equality.*
**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*} n is 3.**

\begin{align*}2x-5=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*} x alone.** Let’s start with the term that is not connected to the \begin{align*}x\end{align*}

We use the ** Addition Property of Equality** to cancel out the -5.

**We do this by using the inverse operation and by 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*}2x-5+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*} x 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*} x is 8.**

\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*} x5 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*}x5, by itself on one side of the equal (=) sign.**

In the equation, 8 is *subtracted* from \begin{align*}\frac{x}{5}\end{align*}*both* sides of the equation because of the ** Addition Property of Equality.** That property states that in order to keep the values on both sides of the equation equal, 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*} x5, is by itself on one side of the equation.**

**We can now use inverse operations to get the \begin{align*}x\end{align*} x by itself. Since \begin{align*}\frac{x}{5}\end{align*}x5 means \begin{align*}x \div 5\end{align*}x÷5, 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

**That property states that if we multiply one side of the equation by a number, we must multiply the other side of the equation by the same number in order to keep the values on both sides of the equation equal.**

*Multiplication Property of Equality.*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*} x is 125.**

Just like when you solved one-step equations, you use properties to solve two-step equations too.

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.

Now it's time for you solve a few on your own.

#### Example A

\begin{align*}5x+7=32\end{align*}

**Solution:\begin{align*}x = 5\end{align*} x=5**

#### Example B

\begin{align*}3a+9=39\end{align*}

**Solution:\begin{align*}a = 10\end{align*} a=10**

#### Example C

\begin{align*}\frac{y}{4}-8=4\end{align*}

**Solution:\begin{align*}y = 3\end{align*} y=3**

Here is the original problem once again.

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 thirty-five 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 two-step equation. Let’s look at what we know.**

**4 tickets at \begin{align*}x\end{align*} x 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+15-15 &= 535-15\\
4x &= 520\\
\frac{4x}{4} &= \frac{520}{4}\\
x &= \$130.00\end{align*}

**Each ticket costs $130.00.**

### Vocabulary

Here are the vocabulary words in this Concept.

- Inverse Operation
- the opposite operation

- One-Step Equation
- an equation with one operation in it

- Two-Step 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.

### Guided Practice

Here is one for you to try on your own.

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*}

b. How many hours did that job take?

**Answer**

Consider part \begin{align*}a\end{align*}

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*}

\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\\ 35-35+20h &= 95-35\\ 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.**

### Video Review

Here is a video for review.

- This is a James Sousa video on solving two -step equations.

### Practice

Directions: Solve each two-step equation for the unknown variable.

1. \begin{align*}3x+2=14\end{align*}

2. \begin{align*}6y+5=29\end{align*}

3. \begin{align*}7x+3=24\end{align*}

4. \begin{align*}5x+7=42\end{align*}

5. \begin{align*}6y+1=43\end{align*}

6. \begin{align*}9a+7=88\end{align*}

7. \begin{align*}11b+12=56\end{align*}

8. \begin{align*}12x-3=21\end{align*}

9. \begin{align*}4y-5=19\end{align*}

10. \begin{align*}3a-9=21\end{align*}

11. \begin{align*}5b-8=37\end{align*}

12. \begin{align*}7x-10=39\end{align*}

13. \begin{align*}6x-12=30\end{align*}

Directions: Write an expression and/or solve each problem.

14. Augusta sells t-shirts at the school store. On Tuesday, Augusta sold 7 less than twice the number of t-shirts she sold on Monday. She sold 3 t-shirts on Tuesday. Write an algebraic equation to represent \begin{align*}m\end{align*}, the number of t-shirts August sold on Monday.

15. 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.

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Here you'll learn to solve two-step equations.