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

### Graphical Solutions to One Variable Inequalities

When you solve a linear inequality, you can represent your solution graphically with a number line. The correct graph will show which of the numbers on the number line are solutions to the inequality.

When graphing solutions to inequalities, use an open circle to identify a number that is *not* included as part of the solution and a closed circle to show that a number *is* included as part of the solution. The line above the number line (or sometimes the shaded section of the number line itself) indicates all of the numbers that are possible solutions to the inequality.

#### Take a look at the solution set to each of the following inequalities on a number line:

- \begin{align*}x+2<5\end{align*}
x+2<5

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

- \begin{align*}2x+6\geq 4.\end{align*}
2x+6≥4.

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

- \begin{align*}-3x + 8 \le 17\end{align*}
−3x+8≤17

*Always** reverse the inequality sign when dividing by a negative number*.

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

### Examples

#### Example 1

Earlier, you were told that 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.

First, write down what you know:

Let \begin{align*}x =\end{align*}

Let \begin{align*}x + 3 =\end{align*}

The equation would therefore be \begin{align*}x + (x+3)>17.\end{align*}

\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})\\
2x &> 14\\
\frac{\cancel{2}x}{{\color{red}\cancel{2}}} &> \frac{\cancel{14}\color{red}{7}}{{\color{red}\cancel{2}}} && (\text{Divide by 2 to solve for the variable})\\
x &> 7\end{align*}

Therefore, Jack's brother must be older than 7. If Jack’s brother is 8 (since \begin{align*}8 > 7\end{align*}

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

#### Example 2

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

\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*}

#### Example 3

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

\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 combine variables})\\
\text{-}3b+4 &\ge 19 && (\text{Simplify})\\
\text{-}3b+4{\color{red}-4} &\ge 19 {\color{red}-4} && (\text{Subtract 4 from both sides})\\
\text{-}3b &\ge 15 && (\text{Simplify})\\
\frac{\text{-}3b}{{\color{red}\text{-}3}} &\le \frac{15}{{\color{red}\text{-}3}} && (\text{Divide by -3, reverse the sign of the inequality})\\
b &\le \text{-}5\end{align*}

#### Example 4

Represent the solution set to the inequality \begin{align*}0.6c + 2 \ge 5.6\end{align*}

\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})\\
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})\\
c &\ge 6\end{align*}

### Review

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

- \begin{align*}\text{-}4v > 12\end{align*}
-4v>12 - \begin{align*}\frac{\text{-}2r}{3} > 4\end{align*}
-2r3>4 - \begin{align*}4(t-2) \le 24\end{align*}
4(t−2)≤24 - \begin{align*}\frac{1}{2} (x+5)>6\end{align*}
12(x+5)>6 - \begin{align*}\frac{1}{4}(g+2)\le 2\end{align*}
14(g+2)≤2 - \begin{align*}0.4(b+2)\ge 2\end{align*}
0.4(b+2)≥2 - \begin{align*}0.5(r-1)<4\end{align*}
0.5(r−1)<4 - \begin{align*}\frac{1}{4}(x+16)>2\end{align*}
14(x+16)>2 - \begin{align*}2-k>5(1-k)\end{align*}
2−k>5(1−k) - \begin{align*}2(1.5c+4)\le \text{-}1\end{align*}
2(1.5c+4)≤-1 - \begin{align*}\text{-}\frac{1}{2}(3x-5)\ge 7\end{align*}
-12(3x−5)≥7 - \begin{align*}0.35+0.10(m-1)< 0.45\end{align*}
0.35+0.10(m−1)<0.45 - \begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)>\frac{1}{2}\end{align*}
14+23(t+1)>12 - The prom committee is selling tickets for a fundraiser for 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.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 2.11.