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# 3.7: Solving Multi-Step Inequalities

Difficulty Level: At Grade Created by: CK-12

## Introduction

Marching Times

“We have a lot to do to get ready for the big parade,” Mrs. Kline announced on Monday.

And Mrs. Kline was definitely correct. The students in the band had learned three new routines that they didn’t quite have down yet. It was going to take extra marching practice to put all of the pieces together. To do this, Mrs. Kline had the students outside marching for extra time.

On Tuesday, the band spent thirty fewer minutes marching than they did on Thursday. All in all, everyone was tired.

“We marched for more than three hours between Tuesday and today,” Juan said on the ride home on Thursday. His feet could definitely tell how long they had been marching, but all in all the hard work was paying off because the pieces were looking much better.

Given this information, what are three possible times that the band marched on Thursday?

Once again, you will need to know about inequalities to solve this problem. This problem will require you to write and solve an inequality that will have more than one step. This lesson will give you all the information that you need to accomplish this task.

What You Will Learn

In this lesson, you will learn how to complete the following skills.

• Solve inequalities involving combinations of like terms.
• Solve inequalities involving parentheses and the distributive property.
• Solve inequalities involving both distributing and combining.
• Model and solve real-world problems using multi-step inequalities.

Teaching Time

I. Solve Inequalities Involving Combinations of Like Terms

As you already know, some inequalities can be solved in a single step. For example, we could solve \begin{align*}b+4<10\end{align*} in one step—by subtracting 4 from each side.

However, two or more steps may be required to solve some inequalities. Inequalities that need more than one inverse operation to solve them can be called multi-step inequalities.

Let’s start by looking at combining like terms when we solve an inequality.

Example

\begin{align*}4x+3x<21\end{align*}

First, you can see that we have two terms that have the same variable. These are like terms. To solve an inequality with like terms, we will need to combine the like terms and then we can solve the inequality using the methods that we have already learned.

\begin{align*}7x<21\end{align*}

Here we divide both sides of the inequality by 7. Multiplication is the inverse operation of division.

\begin{align*}x<3\end{align*}

Now let’s look at one that is a bit more complicated.

Example

Solve for \begin{align*}n\end{align*}: \begin{align*}7n-8n-3>23\end{align*}.

First, subtract \begin{align*}7n-8n\end{align*} because \begin{align*}7n\end{align*} and \begin{align*}8n\end{align*} are like terms. Remember, you will need the rules for working with positive and negative integers when solving these inequalities.

\begin{align*}7n-8n-3 &> 23\\ 7n+(-8n)-3 &> 23\\ [7n+(-8n)]-3 &> 23\\ -1n-3 &> 23\\ -n-3 &> 23.\end{align*}

The expression on the left side of the inequality, \begin{align*}-n-3\end{align*}, is now in simplest form. The 3 cannot be subtracted from the \begin{align*}-n\end{align*} because they are not like terms.

The next step is to isolate the term with the variable, \begin{align*}-n\end{align*}, on one side of the inequality. Since 3 is subtracted from \begin{align*}-n\end{align*}, we should add 3 to both sides of the inequality to do this.

\begin{align*}-n-3 &> 23\\ -n-3+3 &> 23+3\\ -n+(-3+3) &> 26\\ -n+0 &> 26\\ -n &> 26\end{align*}

Since \begin{align*}-n\end{align*} means \begin{align*}-1n\end{align*} or \begin{align*}-1 \times n\end{align*}, we can divide each side of the inequality by -1 to get a positive \begin{align*}n\end{align*} by itself on one side of the equation. Since that involves dividing both sides of the inequality by a negative number, we must reverse the inequality symbol.

\begin{align*}-n &> 26\\ -1n &> 26\\ \frac{-1n}{-1} &< \frac{26}{-1}\\ 1n &< -26\\ n &< -26\end{align*}

The solution for this inequality is \begin{align*}n < -26\end{align*}.

Notice that you performed several operations when solving this inequality. Just as we could solve two step equations, we can also solve two step inequalities.

Let’s look at an example of a two – step inequality.

Example

Solve for \begin{align*}b\end{align*}: \begin{align*}3b+4<10\end{align*}.

Notice that there are two terms on the left side of the inequality, \begin{align*}3b\end{align*} and 4. Our first step should be to use inverse operations to get the term that includes a variable, \begin{align*}3b\end{align*}, by itself on one side of the inequality.

In the inequality, 4 is added to \begin{align*}3b\end{align*}. So, we can use the inverse of addition—subtraction. We can subtract 4 from both sides of the inequality. We do not need to change the inequality symbol during this step because we are subtracting a number, not multiplying or dividing by a negative number.

\begin{align*}3b+4 &< 10\\ 3b+4-4 &< 10-4\\ 3b+0 &< 6\\ 3b &< 6\end{align*}

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

We can now use inverse operations to get the \begin{align*}b\end{align*} by itself. Since \begin{align*}3b\end{align*} means \begin{align*}3 \times b\end{align*}, we can divide both sides of the inequality by 3 to isolate the variable. Since we are dividing by a positive number, not a negative number, the inequality symbol should not change.

\begin{align*}3b &< 6\\ \frac{3b}{3} &< \frac{6}{3}\\ 1b &< 2\\ b &< 2\end{align*}

The solution is \begin{align*}b<2\end{align*}.

Now let’s look at solving another multi-step equation.

Example

Solve for \begin{align*}k\end{align*}: \begin{align*}3k+20+2k+10 \le 5\end{align*}.

First, combine the like terms on the left side of the inequality. \begin{align*}3k\end{align*} and \begin{align*}2k\end{align*} are like terms since each has the same variable, \begin{align*}k\end{align*}. The numbers 20 and 10 are also like terms, so they can be combined as well.

Use the commutative property of addition to help you reorder the terms being added. This property states that terms can be added in any order. Then use the associative property to group the terms so like terms are being added. The associative property of addition states that the grouping of terms being added does not matter. Once you reordered the terms, combine the like terms.

\begin{align*}3k+20+2k+10 & \le 5\\ 3k+(20+2k)+10 & \le 5\\ 3k+(2k+20)+10 & \le 5\\ (3k+2k)+(20+10) & \le 5\\ 5k + 30 & \le 5\end{align*}

The next step is to get the term with the variable, \begin{align*}5k\end{align*}, on one side of the inequality. Since 30 is added to \begin{align*}5k\end{align*}, we should subtract 30 from both sides of the inequality to do this.

\begin{align*}5k+30 & \le 5\\ 5k+30-30 & \le 5-30\\ 5k+0 & \le 5+(-30)\\ 5k & \le -25\end{align*}

Since \begin{align*}5k\end{align*} means \begin{align*}5 \times k\end{align*}, we should divide each side of the inequality by 5—a positive number—to get the \begin{align*}k\end{align*} by itself on one side of the inequality. Be careful! It is true that you will need to divide -25 by 5 to find the solution. However, since you are not dividing both sides of the inequality by a negative number, you do not reverse the inequality symbol.

\begin{align*}5k & \le -25\\ \frac{5k}{5} & \le \frac{-25}{5}\\ 1k & \le -5\\ k & \le -5\end{align*}

The solution is \begin{align*}k \le -5\end{align*} .

Now let’s look at some other types of inequalities.

II. Solve Inequalities Involving Parentheses and the Distributive Property

The inequalities that you will see in this section involve parentheses. We know from our earlier work that we can simplify an equation with parentheses by using the distributive property. We can do this with inequalities as well. Using the distributive property can help you to simplify an inequality so that it is easier to solve.

Let’s look at an example.

Example

Solve for \begin{align*}q\end{align*}: \begin{align*}-9(q+3)<45\end{align*}

Apply the distributive property to the left side of the inequality. Multiply each of the two numbers inside the parentheses by -9 and then add those products.

\begin{align*}-9(q+3) &< 45\\ (-9 \times q)+(-9 \times 3) &< 45\\ -9q+(-27) &< 45\end{align*}

Now, solve as you would solve any two-step inequality. Since -27 is added to \begin{align*}-9q\end{align*}, we can get \begin{align*}-9q\end{align*} by itself on one side of the inequality by subtracting -27 from both sides. Remember, subtracting -27 from a number is the same as adding its opposite, 27, to that number.

\begin{align*}-9q+(-27) &< 45\\ -9q+(-27)-(-27) &< 45-(-27)\\ -9q+(-27+27) &< 45+27\\ -9q+0 &< 72\\ -9q &< 72\end{align*}

To get \begin{align*}q\end{align*} by itself on one side of the inequality, we need to divide both sides by -9. Since we are dividing both sides by a negative number, you need to reverse the inequality symbol.

\begin{align*}-9q &< 72\\ \frac{-9q}{-9} &> \frac{72}{-9}\\ 1q &> -8\\ q &> -8\end{align*}

The solution is \begin{align*}q > -8\end{align*}.

Yes, there are a lot of steps to keep track of, but if you follow each part and don’t skip steps you will get there. Remember, you are trying to solve the inequality. Keeping your focus on this will help you keep from getting confused.

Example

\begin{align*}\frac{1}{2} (x+4) \le 10\end{align*}

First, we use the distributive property to multiply one-half with both of the terms inside the parentheses.

\begin{align*}\frac{1}{2} x+2 \le 10\end{align*}

Next, we subtract two from both sides of the inequality.

\begin{align*}\frac{1}{2} x \le 8\end{align*}

Now we can multiply both sides by the reciprocal of one-half which will cancel out the one-half leaving our variable alone. This is an example of the multiplicative inverse property.

\begin{align*}\frac{2}{1} \left(\frac{1}{2}\right) x \le 8 \left(\frac{2}{1}\right)\end{align*}

The answer is that \begin{align*}x \le 16\end{align*}.

Some inequalities will require that you use combining like terms and the distributive property. Let’s take a look at these problems now.

III. Solve Inequalities Involving Both Distributing and Combining

Now let’s apply both combining like terms and the distributive property when working with the same inequality.

Example

Solve for \begin{align*}w\end{align*}: \begin{align*}-2(8+w)+18<28\end{align*}.

First, we should apply the distributive property to the left side of the inequality. We can multiply each of the two numbers inside the parentheses by -2 and then add those products.

\begin{align*}-2(8+w)+18 &< 28\\ (-2 \times 8)+(-2 \times w)+18 &< 28\\ -16+(-2w)+18 &< 28\end{align*}

Next, we can add the like terms (-16 and 18) on the left side of the inequality. Using the commutative and associative properties to reorder the terms on the left side of the equation can make it easier to see how to do this.

\begin{align*}-16+(-2w)+18 &< 28\\ -16+[(-2w)+18] &< 28\\ -16+[18+(-2w)] &< 28\\ (-16+18)+(-2w) &< 28\\ 2+(-2w) &< 28\end{align*}

Finally, we solve as you would solve any two-step inequality. Since 2 is added to \begin{align*}-2w\end{align*}, our first step should be to subtract 2 from both sides of the inequality.

\begin{align*}2+(-2w) &< 28\\ 2-2+(-2w) &< 28-2\\ 0+(-2w) &< 26\\ -2w &< 26\end{align*}

Now, we can isolate the variable, \begin{align*}w\end{align*}, by dividing both sides of the inequality by -2. Since we are dividing both sides by a negative number, we need to reverse the inequality symbol.

\begin{align*}-2w &< 26\\ \frac{-2w}{-2} &> \frac{26}{-2}\\ 1w &> -13\\ w &> -13\end{align*}

The solution is \begin{align*}w > -13\end{align*}.

All of these skills can help us when we work with real – world problems. Let’s look at solving real – world problems with multi – step inequalities.

IV. Model and Solve Real – World Problems Using Multi-Step Inequalities

Sometimes, we can write a multi-step inequality to represent a real-world problem situation. Once again, you will be using key words to work on these. Take a minute and refer back to your notebook. Then continue.

Now let’s look at an example.

Example

Ms. Layne wants to build a rectangular deck in her back yard. She wants the length of the deck to be exactly 9 feet. She wants the perimeter of her deck to be, at most, 28 feet. The perimeter of any rectangle can be found by using the expression \begin{align*}P=2(l+w)\end{align*}, where \begin{align*}l\end{align*} represents the length and \begin{align*}w\end{align*} represents the width.

a. Write an inequality that could be used to represent \begin{align*}w\end{align*}, the possible widths, in feet, she could use for her deck.

b. Would a deck with a width of 6 feet result in a deck with the perimeter she wants?

Consider part a first.

You know that the length is 9 feet, so substitute 9 for \begin{align*}l\end{align*} into the expression \begin{align*}2(l+w)\end{align*}. This expression represents the actual perimeter of the deck.

\begin{align*}\text{actual perimeter}=2(l+w)=2(9+w)\end{align*}

Since she wants the perimeter to be “at most” 28 feet, you should use the “less than or equal to” \begin{align*}(\le)\end{align*} symbol. Translate this problem into an inequality.

\begin{align*}& She \ wants \ the \ \underline{perimeter} \ of \ her \ deck \ to \ be, \ \underline{at \ most}, \ \underline{28 \ feet}.\\ & \qquad \qquad \qquad \quad \ \ \downarrow \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \quad \ \downarrow\\ & \qquad \qquad \qquad \ 2(9+w) \qquad \qquad \qquad \qquad \quad \ \ \le \qquad \quad 28\end{align*}

So, this problem can be represented by the inequality \begin{align*}2(9+w) \le 28\end{align*}.

Next, consider part b.

To find all the possible values of \begin{align*}w\end{align*}, solve the inequality. First, apply the distributive property to the right side.

\begin{align*}2(9+w) & \le 28\\ (2 \times 9)+(2 \times w) & \le 28\\ 18+2w & \le 28\end{align*}

Now, solve as you would solve any two-step inequality. First, subtract 18 from both sides of the inequality.

\begin{align*}18+2w & \le 28\\ 18-18+2w & \le 28-18\\ 0+2w & \le 10\\ 2w & \le 10\end{align*}

Next, divide both sides of the inequality by 2. Since you are dividing by a positive number, the inequality symbol should stay the same.

\begin{align*}2w & \le 10\\ \frac{2w}{2} & \le \frac{10}{2}\\ 1w & \le 5\\ w & \le 5\end{align*}

The value of \begin{align*}w\end{align*} must be less than or equal to 5. Since 6 is greater than, not less than, 5, it is not a possible value of \begin{align*}w\end{align*}. So, if she built her deck so it was 6 feet wide, it would have a larger perimeter than she wants.

Now let’s look at the problem from the introduction. We can use what we have learned to help us in writing and solving an inequality.

## Real-Life Example Completed

Marching Times

Here is the original problem once again. First, reread it. Then write an inequality and solve it for three possible times for Thursday’s practice. There are four parts to your answer.

“We have a lot to do to get ready for the big parade,” Mrs. Kline announced on Monday.

And Mrs. Kline was definitely correct. The students in the band had learned three new routines that they didn’t quite have down yet. It was going to take extra marching practice to put all of the pieces together. To do this, Mrs. Kline had the students outside marching for extra time.

On Tuesday, the band spent thirty fewer minutes marching than they did on Thursday. All in all, everyone was tired.

“We marched for more than three hours between Tuesday and today,” Juan said on the ride home on Thursday. His feet could definitely tell how long they had been marching, but all in all the hard work was paying off because the pieces were looking much better.

Given this information, what are three possible times that the band marched on Thursday?

Solution to Real – Life Example

First, write an inequality using the given information.

\begin{align*}m - 30\end{align*} is the time that the students practiced on Tuesday

\begin{align*}m\end{align*} is the time that they practiced on Thursday

\begin{align*} >3 \end{align*} hours is the time that they practiced in all.

Here is the inequality.

\begin{align*}m + m - 30 > 3\end{align*} hours or 180 minutes

It makes sense to work with minutes because the given times are in minutes.

Now we solve the inequality.

\begin{align*}2m - 30 &> 180\\ 2m &> 210\\ m &> 105 \ minutes\end{align*}

The band marched for more than 105 minutes on Thursday. You could assume that they marched for 110 minutes, 115 minutes or 120 minutes. There are many possible options.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Inequality
a mathematical statement where one quantity can be less than, greater than or equal to another quantity.
Inverse Operation
the opposite operation. Inverse operations are used to solve equations.
Like Terms
terms in an equation or inequality that have common variables or no variables.
Distributive Property
a term outside of a set of parentheses can be multiplied by each of the terms inside the parentheses. This simplifies the parentheses.

## Time to Practice

Directions: Solve each inequality.

1. \begin{align*}2x+5>13\end{align*}
2. \begin{align*}4x-2<10\end{align*}
3. \begin{align*}6y+9>69\end{align*}
4. \begin{align*}2x-3 \le -4\end{align*}
5. \begin{align*}5x+2 \ge -8\end{align*}
6. \begin{align*}2x-9 \le -5\end{align*}
7. \begin{align*}\frac{x}{3}+1>5\end{align*}
8. \begin{align*}\frac{x}{2}-1<-3\end{align*}
9. \begin{align*}\frac{x}{5}+3>-9\end{align*}
10. \begin{align*}\frac{x}{2}-5>-10\end{align*}
11. \begin{align*}6k-3 > 15\end{align*}
12. \begin{align*}11 - \frac{x}{4} \le 12\end{align*}
13. \begin{align*}12+9j+j<72\end{align*}
14. \begin{align*}12b-3b+5 \ge -31\end{align*}
15. \begin{align*}18+7n+3+6n \le 86\end{align*}
16. \begin{align*}3z-15z-30>54\end{align*}
17. \begin{align*}-2(10+h)<-40\end{align*}
18. \begin{align*}\frac{1}{3}(15+t) \le 26\end{align*}
19. \begin{align*}8(p-3)-2 \ge 14\end{align*}
20. \begin{align*}-3(2+4y)+y>16\end{align*}

Directions: Use inequalities to solve each problem.

Lori went hiking last weekend. On Sunday, she hiked 8 more kilometers than she hiked on Saturday. The total number of kilometers she hiked on both days was less than 40 kilometers.

1. Let \begin{align*}s\end{align*} represent the number of kilometers she hiked on Saturday. Write an inequality to represent this problem.
2. Find three possible values of \begin{align*}s\end{align*}.

Mr. Chang wants to use a rectangular section of his backyard for a vegetable garden. He wants the width of the garden to be exactly 7 feet. He wants the perimeter of the rectangular garden to be no less than 36 feet. The perimeter of any rectangle can be found by using the formula \begin{align*}P=2(l+w)\end{align*}, where \begin{align*}l\end{align*} represents the length and \begin{align*}w\end{align*} represents the width.

1. Write an inequality that could be used to represent \begin{align*}l\end{align*}, the possible lengths, in feet, Mr. Chang could use for his rectangular garden.
2. Would a rectangular garden with a length of 12 feet result in a garden with the perimeter he wants?

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