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# 3.16: Solve Inequalities Involving Combining Like Terms

Difficulty Level: At Grade Created by: CK-12
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Practice Multi-Step Inequalities

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Your community center is selling boxes of cards. Your goal is to sell at least $500 worth of cards. Each box sells for$5 but you also need to pay the company 50 to ship all of the cards to your house to begin selling. Determine the least number of boxes of cards you must sell in order to meet your goal. In this concept, you will learn to solve inequalities involving combining like terms. ### Multi-Step Inequalities You can solve inequalities in many ways. Some inequalities can be solved in a single step. You 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 you solve an inequality. \begin{align*}4x + 3x < 21\end{align*} First, you can see that you have two terms that have the same variable. These are like terms. To solve an inequality with like terms, you will need to combine the like terms and then you can solve the inequality. \begin{align*}7x < 21\end{align*} Here you divide both sides of the inequality by 7. Multiplication is the inverse operation of division. \begin{align*}\begin{array}{rcl} 7x &<& 21 \\ \frac{7x}{7} &<&\frac{21}{7}\\ x&<& 3 \end{array}\end{align*} \begin{align*}x<3\end{align*}The answer is . Let’s look at another example. Solve for \begin{align*}b\end{align*}: \begin{align*}3b + 4 < 10\end{align*}. First, notice that there are two terms on the left side of the inequality, \begin{align*}3b\end{align*} and 4. Therefore, 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, you can use the inverse of addition—subtraction. Therefore subtract 4 from both sides of the inequality. \begin{align*}\begin{array}{rcl} 3b + 4 &<& 10 \\ 3b + 4 − 4 &<& 10 − 4\\ 3b + 0 &<& 6 \\ 3b &<& 6 \end{array}\end{align*}Next, 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*}, divide both sides of the inequality by 3 to solve for the variable \begin{align*}\begin{array}{rcl} 3b &<& 6 \\ \frac{3b}{3} &<&\frac{6}{3}\\ b&<& 2 \end{array}\end{align*} The answer is \begin{align*}b < 2\end{align*}. ### Examples #### Example 1 Earlier, you were given a problem about boxes of cards being sold. First, write an inequality using the given information. You want to raise at least500 by selling boxes of cards worth $5 each. You also need to pay the supplier$50 for shipping.  Let ‘\begin{align*}b\end{align*}’ represent the number of boxes of cards.

\begin{align*}5b-50 \ge 500\end{align*}

Next, isolate the term with the variable, \begin{align*}5b\end{align*}, on one side of the inequality. Since 50 is subtracted from \begin{align*}5b\end{align*}, you should add 50 to both sides of the inequality.

\begin{align*}\begin{array}{rcl} 5b-50&\ge& 500 \\ 5b-50+50 &\ge& 500+50\\ 5b&\ge& 550 \end{array}\end{align*}

Next, since \begin{align*}5b\end{align*} means \begin{align*}5 \times b\end{align*}, you can divide each side of the inequality by 5 to get ‘\begin{align*}b\end{align*}’ by itself on one side of the inequality.

\begin{align*}\begin{array}{rcl} 5b &\ge& 550 \\ \frac{5b}{5} &\ge& \frac{550}{5}\\ b&\ge& 110 \end{array} \end{align*}

The answer is \begin{align*}b \ge 110\end{align*}.

You need to sell at least 110 boxes of cards.

#### Example 2

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.

\begin{align*}\begin{array}{rcl} 7n− 8n− 3 & >& 23 \\ 7n+(− 8n)− 3 &<& 23 \\ -n-3 &<& 23 \end{array}\end{align*}

Next, 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*}, you should add 3 to both sides of the inequality.

\begin{align*}\begin{array}{rcl} -n-3 &<& 23 \\ -n-3+3 &<& 23+3\\ -n &<& 26 \end{array}\end{align*}

Then, since \begin{align*}-n\end{align*} means \begin{align*}-1n\end{align*} or \begin{align*}-1 \times n\end{align*}, you 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, you must reverse the inequality symbol.

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

The answer is \begin{align*}n >-26\end{align*}.

#### Example 3

Solve for \begin{align*}x\end{align*}:

\begin{align*}4x + 5 < 21\end{align*}

First, isolate the term with the variable, \begin{align*}4x\end{align*}, on one side of the inequality.  Since 5 is added to \begin{align*}4x\end{align*}, you should subtract 5 from both sides of the inequality.

\begin{align*}\begin{array}{rcl} 4x+5 &<& 21 \\ 4x+5-5 &<& 21 -5\\ 4x&<& 16 \end{array}\end{align*}

Next, since \begin{align*}4x\end{align*} means \begin{align*}4 \times x\end{align*}, you can divide each side of the inequality by 4 to get ‘\begin{align*}x\end{align*}’ by itself on one side of the inequality.

\begin{align*}\begin{array}{rcl} 4x &<& 16 \\ \frac{4x}{4} &<& \frac{16}{4}\\ x&<& 4 \end{array}\end{align*}

The answer is \begin{align*}x < 4\end{align*}.

#### Example 4

Solve for \begin{align*}x\end{align*}:

\begin{align*}3x-6>30\end{align*}

First, isolate the term with the variable, \begin{align*}3x\end{align*}, on one side of the inequality. Since 6 is subtracted from \begin{align*}3x\end{align*}, you should add 6 to both sides of the inequality.
\begin{align*}\begin{array}{rcl} 3x -6&>& 30 \\ 3x-6+6 &>& 30+6\\ 3x&>& 36 \end{array}\end{align*}

Next, since \begin{align*}3x\end{align*} means \begin{align*}3 \times x\end{align*}, you can divide each side of the inequality by 3 to get ‘\begin{align*}x\end{align*}’ by itself on one side of the inequality.

\begin{align*}\begin{array}{rcl} 3x &>&36 \\ \frac{3x}{3} &>& \frac{36}{3}\\ x&>& 12 \end{array}\end{align*}

.\begin{align*}x>12\end{align*}The answer is

#### Example 5

Solve for \begin{align*}a\end{align*}

\begin{align*}-3a+2<14\end{align*}

First, isolate the term with the variable, \begin{align*}-3a\end{align*}, on one side of the inequality. Since 2 is added to \begin{align*}-3a\end{align*}, you should subtract 2 from both sides of the inequality.

\begin{align*}\begin{array}{rcl} -3a+2&<& 14 \\ -3a+2-2 &<& 14-2\\ -3a&<& 12 \end{array}\end{align*}

Next, since\begin{align*}-3a\end{align*} means \begin{align*}-3 × a\end{align*}, you can divide each side of the inequality by -3 to get ‘\begin{align*}a\end{align*}’ by itself on one side of the equation. Remember to reverse the inequality sign since you are dividing by a negative number.

\begin{align*}\begin{array}{rcl} -3a &<& 12 \\ \frac{-3a}{-3} &<& \frac{12}{-3}\\ a&>& -4 \end{array}\end{align*} The answer is \begin{align*}a > -4\end{align*}.

### Review

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*}

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### Vocabulary Language: English

TermDefinition
inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
Inverse Operation Inverse operations are operations that "undo" each other. Multiplication is the inverse operation of division. Addition is the inverse operation of subtraction.
like terms Terms are considered like terms if they are composed of the same variables with the same exponents on each variable.

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