The average weight gain of an infant, after 6 months of age, is one pound a month, until the age of 2. If the average 6-month-old weighs 16 pounds, up to what age would an infant weigh 25 pounds or less?

### Basic Inequalities

Solving a linear inequality is very similar to solving a linear equality, or equation. There are a few very important differences. We no longer use an equal sign. There are four different inequality signs, shown below.

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

\begin{align*} > \end{align*}

\begin{align*} \le \end{align*}

\begin{align*} \ge \end{align*}

Notice that the line underneath the \begin{align*}\le\end{align*}

Notice that the circle at -1 is *open*. This is to indicate that -1 is *not* included in the solution. A < sign would also have an open circle. If the inequality was a \begin{align*}\ge\end{align*}

Let's determine whether \begin{align*}x=-8\end{align*}

Plug in -8 for \begin{align*}x\end{align*}

\begin{align*}\frac{1}{2}(-8)+6 &> 3\\
-4+6 &> 3\\
2 &> 3\end{align*}

Of course, 2 cannot be greater than 3. Therefore, this is not a valid solution.

Now, let's solve the following basic inequalities.

- Solve and graph the solution to \begin{align*}2x-5 \le 17\end{align*}
2x−5≤17 .

For the most part, solving an inequality is the same as solving an equation. The major difference will be addressed in problem #2 below. This inequality can be solved just like an equation.

\begin{align*}& 2x-\bcancel{5} \le 17\\
& \underline{\quad \ + \bcancel{5} \ +5 \; \;}\\
& \quad \ \frac{\bcancel{2}x}{\bcancel{2}} \le \frac{22}{2}\\
& \qquad x \le 2\end{align*}

Test a solution, \begin{align*}x = 0: 2(0)-5 \le 17 \rightarrow -5 \le 17 \checkmark\end{align*}

Plotting the solution, we get:

Always test a solution that is in the solution range. It will help you determine if you solved the problem correctly.

- Solve and graph \begin{align*}-6x+7 \le - 29\end{align*}.

When solving inequalities, be careful with negatives. Let’s solve this problem the way we normally would an equation.

\begin{align*}& -6x+\bcancel{7} \le -29\\ & \underline{\qquad \ \ -\bcancel{7} \quad \ -7 \; \; \; \;}\\ & \qquad \frac{-\bcancel{6}x}{-\bcancel{6}x} \le \frac{-36}{-6}\\ & \qquad \quad \ x \le 6\end{align*}

Let’s check a solution. If \begin{align*}x\end{align*} is less than or equal to 6, let’s test 1.

\begin{align*}-6(1)+7 & \le -29\\ -6+7 & \le -29\\ 1 & \bcancel{\le} -29\end{align*}

This is not a true inequality. To make this true, we must *flip* the inequality. Therefore, **whenever we multiply or divide by a negative number, we must flip the inequality sign.** The answer to this inequality is actually \begin{align*}x\ge 6\end{align*}. Now, let’s test a number in this range.

\begin{align*}-6(10)+7 & \le - 29\\ -60+7 & \le -29\\ -60 & \le -29\end{align*}

This is true. The graph of the solution is:

### Examples

#### Example 1

Earlier, you were asked to find the last age (the maximum age) that an infant would weigh 25 pounds or less.

First, write an inequality. Let *m* represent the age of the infant, in months. Remember, when you get the final answer, you must add 6 for the initial weight of the infant at 6 months.

\begin{align*}16 + m \le 25 \\ m \le 9 \end{align*}

Adding 6, we have \begin{align*}m \le 15\end{align*}. So, up to 15 months, the average baby should weigh 25 pounds or less.

#### Example 2

Is \begin{align*}x = -5\end{align*} a solution to \begin{align*}-3x+7>12\end{align*}?

Plug -5 into the inequality.

\begin{align*}-3(-5)+7 &>12\\ 15+7&>12\end{align*}

This is true because 22 is larger than 12. -5 is a solution.

#### Example 3

Solve and graph the solution to \begin{align*}\frac{3}{8}x+5<26\end{align*}.

No negatives with the \begin{align*}x-\end{align*}term, so we can solve this inequality like an equation.

\begin{align*}& \frac{3}{8}x+\bcancel{5}<26\\ & \underline{\quad \ -\bcancel{5} \ \ -5 \; \; \; \; \; \;}\\ & \ \xcancel{\frac{8}{3} \cdot \frac{3}{8}}x<21 \cdot \frac{8}{3}\\ & \qquad \ x<56\end{align*}

Test a solution, \begin{align*}x = 16: \frac{3}{8}(16)+5 < 26 \checkmark\end{align*}

\begin{align*}6+5 < 26\end{align*}

The graph looks like:

#### Example 4

Solve and graph the solution to \begin{align*}11<4-x\end{align*}.

In this inequality, we have a negative \begin{align*}x-\end{align*}term. Therefore, we will need to flip the inequality.

\begin{align*}& \ \ 11<\bcancel{4}-x\\ & \underline{-4 \ \ - \bcancel{4} \; \; \; \; \; \; \;}\\ & \frac{7}{-1} < \frac{\bcancel{-}x}{\bcancel{-1}}\\ & -7>x\end{align*}

Test a solution, \begin{align*}x = -10: 11 < 4-(-10) \checkmark\end{align*}

\begin{align*}11 < 14\end{align*}

Notice that we *flipped* the inequality sign when we *divided* by -1. Also, this equation can also be written \begin{align*}x< -7\end{align*}.

Here is the graph:

### Review

Solve and graph each inequality.

- \begin{align*}x+5>-6\end{align*}
- \begin{align*}2x \ge 14\end{align*}
- \begin{align*}4<-x\end{align*}
- \begin{align*}3x-4 \le 8\end{align*}
- \begin{align*}21-8x<45\end{align*}
- \begin{align*}9>x-2\end{align*}
- \begin{align*}\frac{1}{2}x+5 \ge 12\end{align*}
- \begin{align*}54 \le -9x\end{align*}
- \begin{align*}-7<8+\frac{5}{6}x\end{align*}
- \begin{align*}10-\frac{3}{4}x<-8\end{align*}
- \begin{align*}4x+15 \ge 47\end{align*}
- \begin{align*}0.6x-2.4<4.8\end{align*}
- \begin{align*}1.5>-2.7-0.3x\end{align*}
- \begin{align*}-11<12x+121\end{align*}
- \begin{align*}\frac{1}{2}-\frac{3}{4}x \le -\frac{5}{8}\end{align*}

For questions 16 and 17, write the inequality statement given by the graph below.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 1.10.