Jack is 3 years older than his brother. How old are they if the sum of their ages is greater than 17? Write an inequality and solve. Represent the solution set on a number line.

### Watch This

Khan Academy Inequalities on a Number Line

### Guidance

When you solve a linear inequality you can represent your solution graphically with a number line. Your job is to show which of the numbers on the number line are solutions to the inequality.

When graphing solutions to inequalities, use an open circle to show that the number is not included as part of the solution and a closed circle to show that the number is included as part of the solution. The line above the number line shows all of the numbers that are possible solutions to the inequality.

#### Example A

Represent the solution set to the following inequality on a number line: \begin{align*}x+2<5\end{align*}.

**Solution:**

\begin{align*}x + 2 &< 5\\ x+2{\color{red}+-} &< 5{\color{red}-22} && (\text{Subtract 3 from both sides to isolate the variable})\\ x &< 3 && (\text{Simplify})\end{align*}

#### Example B

Represent the solution set to the following inequality on a number line: \begin{align*}2x+6\geq 4\end{align*}.

**Solution:**

\begin{align*}2x + 6 &\ge 4\\ 2x+6{\color{red}-6} &\ge 4{\color{red}-6} && (\text{Subtract 6 from both sides to isolate the variable})\\ 2x &\ge -2 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &\ge \frac{-2}{{\color{red}2}} && (\text{Divide by 2 to solve for the variable})\\ x &\ge -1\end{align*}

#### Example C

Represent the solution set to the following inequality on a number line: \begin{align*}-3x + 8 \le 17\end{align*}.

**Solution:**

\begin{align*}-3x + 8 &\le 17\\ -3x+8{\color{red}-8} &\le 17{\color{red}-8} && (\text{Subtract 8 from both sides to isolate the variable})\\ -3x &\le 9 && (\text{Simplify})\\ \frac{-3x}{{\color{red}-3}} &\ge \frac{9}{{\color{red}-3}} && (\text{Divide both sides by -3 to solve for the variable, reverse sign of inequality})\\ x &\ge -3\end{align*}

#### Concept Problem Revisited

Jack is 3 years older than his brother. How old are they if the sum of their ages is greater than 17? Write an inequality and solve. Represent the solution set on a number line.

If first let’s write down what you know:

Let \begin{align*}x =\end{align*} Jack’s brother’s age

Let \begin{align*}x + 3 =\end{align*} Jack’s age

The equation would therefore be:

\begin{align*}x+x+3 &> 17\\ 2x+3 &> 17 && (\text{Combine like terms})\\ 2x+3{\color{red}-3} &> 17{\color{red}-3} && (\text{Subtract 3 from both sides to solve for the variable})\\ 2x &> 14 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &> \frac{14}{{\color{red}2}} && (\text{Divide by 2 to solve for the variable})\\ x &> 7\end{align*}

Therefore if Jack’s brother is 8 (since \begin{align*}8 > 7\end{align*}), Jack would be 11. If Jack's brother is 10, Jack would be 13.

Representing Jack's brother's age on a number line:

### Vocabulary

- Number Line
- A
is a line that matches a set of points and a set of numbers one to one.*number line*

### Guided Practice

1. Represent the solution set to the inequality \begin{align*}3(a - 1) < 9\end{align*} on a number line.

2. Represent the solution set to the inequality \begin{align*}2b + 4 \ge 5b + 19\end{align*} on a number line.

3. Represent the solution set to the inequality \begin{align*}0.6c + 2 \ge 5.6\end{align*} on a number line.

**Answers:**

1.

\begin{align*}3(a-1)&<9\\ 3a-3 &< 9 && (\text{Remove parentheses})\\ 3a-3{\color{red}+3} &< 9 {\color{red}+3} && (\text{Add 3 to both sides to isolate the variable})\\ 3a &< 12 && (\text{Simplify})\\ \frac{3a}{{\color{red}3}} &< \frac{12}{{\color{red}3}} && (\text{Divide both sides by 3 to solve for the variable})\\ a &< 4\end{align*}

2.

\begin{align*}2b+4 &\ge 5b+19\\ 2b{\color{red}-5b}+4 &\ge 5b{\color{red}-5b}+19 && (\text{Subtract} \ 5b \ \text{from both sides to get variables on same side})\\ -3b+4 &\ge 19 && (\text{Simplify})\\ -3b+4{\color{red}-4} &\ge 19 {\color{red}-4} && (\text{Subtract 4 from both sides to isolate the variable})\\ -3b &\ge 15 && (\text{Simplify})\\ \frac{-3b}{{\color{red}-3}} &\le \frac{15}{{\color{red}-3}} && (\text{Divide by -3 to solve for the variable, reverse sign of inequality})\\ b &\le -5\end{align*}

3.

\begin{align*}0.6 c+2 &\ge 5.6\\ 0.6c+2{\color{red}-2} &\ge 5.6{\color{red}-2} && (\text{Subtract 2 from both sides to isolate the variable})\\ 0.6c &\ge 3.6 && (\text{Simplify})\\ \frac{0.6c}{{\color{red}0.6}} &\ge \frac{3.6}{{\color{red}0.6}} && (\text{Divide both sides by 0.6 to solve for the variable})\\ c &\ge 6\end{align*}

### Practice

Solve each inequality and represent the solution set on a number line.

- \begin{align*}-4v > 12\end{align*}
- \begin{align*}\frac{-2r}{3} > 4\end{align*}
- \begin{align*}4(t-2) \le 24\end{align*}
- \begin{align*}\frac{1}{2} (x+5)>6\end{align*}
- \begin{align*}\frac{1}{4}(g+2)\le 2\end{align*}
- \begin{align*}0.4(b+2)\ge 2\end{align*}
- \begin{align*}0.5(r-1)<4\end{align*}
- \begin{align*}\frac{1}{4}(x+16)>2\end{align*}
- \begin{align*}2-k>5(1-k)\end{align*}
- \begin{align*}2(1.5c+4)\le -1\end{align*}
- \begin{align*}-\frac{1}{2}(3x-5)\ge 7\end{align*}
- \begin{align*}0.35+0.10(m-1)< 0.45\end{align*}
- \begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)>\frac{1}{2}\end{align*}
- The prom committee is selling tickets for a fundraiser for the decorations. Each ticket costs $3.50. What is the least number of tickets the committee needs to sell to make $1000? Write an inequality and solve.
- Brenda got 69%, 72%, 81%, and 88% on her last four major tests. How much does she need on her next test to have an average of at least 80%? Write an inequality and solve.

### Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 2.11.