# 2.3.1: Two-Step Equations

- Page ID
- 4371

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\id}{\mathrm{id}}\)

\( \newcommand{\Span}{\mathrm{span}}\)

\( \newcommand{\kernel}{\mathrm{null}\,}\)

\( \newcommand{\range}{\mathrm{range}\,}\)

\( \newcommand{\RealPart}{\mathrm{Re}}\)

\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

\( \newcommand{\Argument}{\mathrm{Arg}}\)

\( \newcommand{\norm}[1]{\| #1 \|}\)

\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)

\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)

\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vectorC}[1]{\textbf{#1}} \)

\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## Two-Step Equations and Properties of Equality

As a mid-term project in a music class, some students have the opportunity to see a live jazz concert and describe the music to the class. The music teacher, Mr. Cooper, purchased some tickets. The service fee to buy tickets to the jazz concert is 9 dollars, and each ticket costs 12 dollars. If the total cost of the tickets was 93 dollars, can you figure out how many tickets Mr. Cooper his bought?

In this concept, you will learn to solve two-step equations.

**Solving Two-Step Equations**

Sometimes, when the numbers are small integers, you might be able to solve a **two-step equation** by thinking about it. For instance, can you solve 3x+3=9?

Here the numbers are small. You can probably look at this equation and ask yourself, “What number times three plus three is nine?” The logical answer is 2. You can check your answer by substituting 2 in for x, to see if both sides of the equation are the same. If they are, then your work is accurate.

Let’s plug in 2 for x to check that possible answer.

3(2)+3=9

6+3=9

9=9

Since it is true that 9=9, the answer x=2 works.

As equations get more complex it is important to use properties of equality to **isolate the variable** and solve the equation.

Here are the properties of equality you need to isolate terms and solve equations.

The **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 the equation without changing the equality.

The **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 the equation without changing the equality.

The **Division Property of Equality** is used when you have an equation with a variable multiplied by a number. It states that you can divide both sides of an equation by the same quantity (as long as that quantity is not equal to zero) without changing the equality.

The **Multiplication Property of Equality** is used when you have an equation with a variable divided by a number. It states that you can multiply both sides of an equation by the same quantity without changing the equality.

Let’s look at an example and use properties of equality to isolate the variable and solve the equation.

2+3n=11

There are two terms on the left side of the equation, 2 and 3n.

The first step is to get the term with the variable, 3n, by itself on one side of the equal (=) sign.

In the equation, 2 is added to 3n. So, you use the inverse of addition, which is subtraction, and subtract 2 from both sides of the equation.

You can do this because of the subtraction property of equality. That property states that in order to keep the values on both sides of the equation equal, whatever is subtracted 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

2+3n=11

2−2+3n=11-2

0+3n=9

3n=9

Now the problem is much more simple. You have reduced a two-step equation to a one-step equation.

Next, use the division property of equality and divide both sides of the equation by 3. Then simplify.

3n=9

3n/3=9/3

1n=3

n=3

The answer is n=3.

Here is another example.

Solve 2x−5=11 for x.

This is a two-step equation. The ultimate goal is to isolate the variable x. This will be easier to do if term with the variable, 2x is by itself on one side of the equation.

First, to get 2x by itself on one side of the equation, use the addition property of equality to add 5 to both sides of the equation.

2x−5=11

2x−5+5=11+5

2x+0=16

2x=16

Now, the two-step equation is a one-step equation and is much easier to solve.

Next, use the division property of equality, and divide both sides of the equation by 2, to isolate the variable x.

2x/2=16

1x=8

x=8

The answer is x=8.

Here is another example.

Solve (x/5)−8=17 for x.

First, use the addition property of equality to get x5 by itself on one side of the equation. Add 8 to both sides of the equation.

(x/5)−8=17

(x/5)+(−8)+8=17+8

(x/5)+0=25

x/5=25

Now, the two-step equation has been reduced to a one-step equation. Since x is divided by 5, you need to use the inverse of division, multiplication, to isolate the variable x.

Next, use the multiplication property of equality, and multiply both sides of the equation by 5. Then simplify both sides of the equation.

(x/5)×5=25x5

x×(5/5)=125

x×1=125

x=125

The answer is x=125.

**Examples**

Example 2.3.1.1

Earlier, you were given a problem about Mr. Cooper and his music class.

He bought tickets to an amazing jazz concert for some students. There was a $9 service fee per order, and each ticket cost $12. The total cost for the order was $93. Can you write an equation to represent this situation and then solve it?

**Solution**

First, let n be the number of tickets bought. The total cost is 9 plus 12 times the number of tickets bought. This total cost is 93. Translate this into an equation.

12n+9=93

The answer is 12n+9=93.

Next, solve the two-step equation.

First isolate 12n. Use the subtraction property of equality and subtract 9 from both sides of the equation.

12n+9=93

12n+9−9=93-9

12n+0=84

12n=84

Now, solve the one-step equation. Use the division property of equality and divide both sides of the equation by 12.

12n/12=84/12

(12/12)n=7

1n=7

n=7

The answer is Mr. Cooper bought 7 tickets.

Example 2.3.1.2

A landscaper charges $35 for each landscaping job, plus $20 for each hour worked. After one landscaping job, the landscaper charged $95.

- Write an algebraic equation to represent h, the number of hours that the landscaper worked on that $95 job.
- How many hours did that job take?

**Solution**

Consider part 1 first.

The landscaper earned $20 for each hour worked on that job, so you multiply $20 by h, the number of hours worked, to find how the landscaper charged for the hours worked, and then add the initial $35. This equals the total charge of $95.

The answer is 20h+35=95.

Next, consider part 2.

Solve the equation to find the number of hours the landscaper worked on that job.

First, to isolate the term 20h use the subtraction property of equality and subtract 35 from both sides of the equation.

20h+35−35=95-35

20h+0=60

20h=60

The two-step equation has been reduced to a one-step equation.

Next, use the division property of equality to isolate the variable h, and divide both sides of the equation by 20.

20h/20=60/20

(20/20)h=3

1h=3

h=3

The answer is the landscaper worked for 3 hours on the $95 job.

**Solve each equation.**

Example 2.3.1.3

5x+7=32

**Solution**

First, use the subtraction property of equality and subtract 7 from both sides of the equation.

5x+7−7=32-7

5x+0=25

5x=25

Next, use the division property of equality and divide each side of the equation by 5.

5x/5=25/5

(5/5)x=5

1x=5

x=5

The answer is x=5.

Example 2.3.1.4

3a+9=39

**Solution**

First, use the subtraction property of equality and subtract 9 from both sides of the equation.

3a+9−9=39-9

3a+0=30

3a=30

Next, use the division property of equality and divide both sides of the equation by 3.

3a/3=30/10

(3/3)a=10

1a=10

a=10

The answer is a=10.

Example 2.3.1.5

(y/4)−8=4

**Solution**

First, use the addition property of equality and add 8 to both sides of the equation.

(y/4)−8+8=4+8

(y/4)+0=12

y/4=12

Next, use the multiplication property of equality and multiply both sides of the equation by 4.

(y/4)×4=12x4

y×(4/4)=48

y×1=48

y=48

The answer is y=48.

**Review**

Solve each two-step equation for the unknown variable.

- 3x+2=14
- 6y+5=29
- 7x+3=24
- 5x+7=42
- 6y+1=43
- 9a+7=88
- 11b+12=56
- 12x−3=21
- 4y−5=19
- 3a−9=21
- 5b−8=37
- 7x−10=39
- 6x−12=30

Write an equation for each word problem and then solve for the unknown variable.

- 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 m, the number of t-shirts August sold on Monday.
- 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 r, the number of red marbles in the box.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.12.

### Vocabulary

Term | Definition |
---|---|

Addition Property of Equality |
The addition property of equality states that you can add the same quantity to both sides of an equation without changing the relative truth of the statement. If 2x=6, then 2x+2=6+2. |

Division Property of Equality |
The division property of equality states that two equal values remain equal if both are divided by the same number. For example: If 2x=8, then 1x=4. |

Inverse Operation |
Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction. |

Isolate the variable |
To isolate the variable means to manipulate an equation so that the indicated variable is by itself on one side of the equals sign. |

Multiplication Property of Equality |
The multiplication property of equality states that if the same constant is multiplied to both sides of the equation, the equality holds true. |

Subtraction Property of Equality |
The subtraction property of equality states that you can subtract the same quantity from both sides of an equation and it will still balance. |

Two-Step Equation |
A two-step equation is an algebraic equation with two operations in it that requires two steps to solve. |

### Additional Resources

Video:

PLIX: Play, Learn, Interact, eXplore: **T-Shirt Equation**

Activity: **Two-Step Equations and Properties of Equality Discussion Questions**

Practice: **Two-Step Equations**

Real World Application: **Calories: Getting a Bad Rap**