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# 3.1: Solve Equations Involving Inverse Properties of Addition and Multiplication

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Practice Two-Step Equations with Addition and Multiplication
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The marching band at Floyd Middle School is excellent and well respected throughout the community. Each year the band continues to grow, and the students who participate make a huge commitment to practices, football games and parades.

When the students gathered for their first rehearsal, Mrs. Kline the band director gathered them altogether for a few announcements.

“We will be marching in the big parade this year once again,” she said smiling from ear to ear.

“I am so glad,” Keri said leaning over to Anica. “I was hoping that we would.”

“We will also be adding four new students this year. We started with 140 students and we will add four. This means that we need to redo our formation for the big finale. We need to reorganize the band into eight even rows. Let’s take a look at what this will look like. Please take out a piece of paper and a pencil,” Mrs. Kline said turning to the blackboard.

“I can figure out the number of students in each row with an equation,” Anica said smiling.

“Yes, and don’t forget to count Jake as the Drum Major in the lead,” Keri added.

Do you know what this equation needs to look like? We have been given the sum of the students, a Drum Major and we know that we need eight even rows. As Anica said, we will need an equation to figure out the number of students in each row. This Concept will teach you all that you need to know about equations so that you will know how to solve this in the end.

### Guidance

What do you know about equations?

An equation is a statement with an equal sign where the quantity on one side of the equals is the same as the quantity on the other side of the equals.

Here is a simple equation.

$x+11=15$

Here we have an equation with a variable where $x$ is the unknown quantity. To solve this, the natural thing to do is to perform an inverse operation or opposite operation . We subtract eleven from 15 which leaves us with 4. That is the value of the variable.

Most of the time, you don’t even think about performing an inverse operation, your mind naturally solves the problem in this way.

When you have an equation with one variable, it is called a one-step equation . It only takes one operation or one inverse operation to solve it. You have had a lot of practice solving one-step equations.

To solve a two-step equation, we will need to use more than one inverse operation.

Let's take a look at how to solve a two-step equation.

When we perform inverse operations to find the value of a variable, we work to get the variable alone on one side of the equals. This is called isolating the variable. It is one strategy for solving equations. You can use isolating the variable whether you are solving one-step or two-step equations.

Here is a two-step equation.

Solve for $a$ : $3a + 12 = 45$ .

We can call each piece of the equation a term . There is a term with a variable and there is a term without a variable. Notice that there are two terms on the left side of the equation, $3a$ and 12.

Our first step is to use inverse operations to get the term that includes a variable, $3a$ , by itself on one side of the equal (=) sign. Because the three is connected to the variable, we perform the other inverse operation first. We work with the number connected with the variable last.

In the equation, 12 is added to $3a$ . So, we can use the inverse of addition—subtraction. We can subtract 12 from both sides of the equation.

Let's see what happens when we subtract 12 from both sides of the equation.

$3a + 12 & = 45\\3a + 12 - 12 & = 45 - 12\\3a + 0 & = 33\\3a & = 33$

Now, the term that includes a variable, $3a$ , is by itself on one side of the equation.

We can now use inverse operations to get the $a$ by itself. Since $3a$ means $3 \times a$ , we can use the inverse of multiplication—division. We can divide both sides of the equation by 3. Let's see what happens when we divide both sides of the equation by 3.

$3a & = 33\\\frac{3a}{3} & = \frac{33}{3}\\1a & = 11\\a & = 11$

The value of $a$ is 11.

Let’s review our steps to solving this two-step equation.

Take a few minutes to write these steps in your notebook.

#### Example A

$4x + 5 = 29$

Solution: $x = 6$

#### Example B

$3y + 7 = 43$

Solution: $y = 12$

#### Example C

$6x + 8 = 71$

Solution: $x = 9$

Now let's go back to the dilemma from the beginning of the Concept.

First, let’s look at the given information.

There are 144 students in the band.

There is also one Drum Major.

We need to organize the students into eight even rows.

Here is our equation.

$8x=144$

Now you may be wondering why we didn’t include the Drum Major. Well, as Keri points out, the Drum Major is in the lead. In this case, Jake is not included in the equation since he is not in the rows.

We have a one-step equation here. We can solve the equation now.

$x=18 \ students$

There will be 18 students in each row.

### Vocabulary

Equation
a mathematical statement with an equal sign where the quantity on one side of the equation is equal to the quantity on the other side.
Variable
a letter used to represent an unknown quantity.
Algebraic Equation
An equation with at least one variable in it.
One-Step Equation
An algebraic equation with one operation in it.
Two-Step Equation
An algebraic equation with two operations in it.

### Guided Practice

Here is one for you to try on your own.

A gardener charges $20 for each gardening job plus$15 for each hour worked. He charged $80 for a gardening job he did yesterday. a. Write an algebraic equation to represent $h$ , the number of hours that the gardener worked on that$80 job.

b. Find the number of hours that the gardener worked on that $80 job. Solution Consider part a first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. The gardener earned$15 for each hour worked on that job, so you could multiply $15 by $h$ , the number of hours worked, to find how much money the gardener charged for his worktime. $& \underline{\ 20} \ for \ each \ gardening \ job \ \underline{plus} \ \underline{\15 \ for \ each \ hour} \ worked \ldots \underline{charged} \ \underline{\80} \ for \ one \ldots job.\\& \downarrow \qquad \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \downarrow \qquad \qquad \qquad \qquad \ \ \downarrow \qquad \downarrow\\& 20 \qquad \qquad \qquad \qquad \qquad \quad \ + \qquad \qquad \quad 15h \qquad \qquad \qquad \qquad = \quad \ 80$ So, this equation, $20 + 15h = 80$ , represents $h$ , the number of hours the gardener worked on the$80 job.

Next, consider part b .

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

Since 20 is added to the term that includes a variable, $15h$ , we can use the inverse of addition—subtraction. We can subtract 20 from both sides of the equation, like this:

$20 + 15h & = 80\\20 - 20 + 15h & = 80 - 20\\0 + 15h & = 60\\15h & = 60$

Since 15 is multiplied by the variable, $h$ , we can use the inverse of multiplication—division. We can divide both sides by 15 to solve for $h$ , like this:

$15h & = 60\\\frac{15h}{15} & = \frac{60}{15}\\1h & = 4\\h & = 4$

The gardener worked four hours on the job he did yesterday.

### Practice

Directions: Solve the following two-step equations that have addition and multiplication in them.

1. $3x + 4 = 22$
2. $4y + 3 = 15$
3. $6x + 5 = 35$
4. $7x + 2 = 16$
5. $9y + 8 = 80$
6. $12x + 15 = 51$
7. $14y + 2 = 30$
8. $7y + 5 = 40$
9. $2x + 4 = 48$
10. $6x + 3 = 39$
11. $8x + 2 = 10$
12. $8x + 7 = 95$
13. $9x + 9 = 90$
14. $3x + 5 = 50$
15. $7x + 12 = 61$

Basic

Dec 19, 2012

Oct 28, 2014