In this concept, you will learn to solve linear inequalities.

### Linear Inequalities

When solving a linear equation with one variable the goal is to determine a value for the variable that makes the equation true. When solving a linear equation with two variables the goal is to determine an ordered pair (\begin{align*}x, y\end{align*}) that makes the equation true. When solving a system of two linear equations with two variables the goal is to determine an ordered pair (\begin{align*}x, y\end{align*}) that makes both equations true. The goals are the same for solving an inequality with one variable, two variables or a system of linear inequalities.

Unlike an equation that has a single solution, an inequality has more than one solution which is indicated by the inequality symbol. An equation has an equals sign but an inequality will have one of the following symbols:

\begin{align*}>\end{align*} which is read as “greater than”

which is read as “greater than and equal to”

which is read as “less than”

which is read as “less than and equal to”

Let’s look at an example.

Solve the following inequality:

\begin{align*}4x+2 < 18\end{align*}

Solve the inequality as if it were a linear equation.

First, subtract 2 from both sides of the inequality and simplify both sides of the equation.

\begin{align*}\begin{array}{rcl} 4x+2 &<& 18\\ 4x+2-2 &<& 18-2\\ 4x &<& 16 \end{array}\end{align*}

Next, divide both sides of the inequality by 4 to solve for the variable ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 4x &<& 16\\ \frac{\overset{1}{\cancel{4}}}{\cancel{4}}x &<& \frac{\overset{4}{\cancel{16}}}{\cancel{4}}\\ x &<& 4 \end{array}\end{align*}

Any value for the variable less than four is a solution for the given inequality. The solution can be expressed as \begin{align*}x<4\end{align*} or as using set notation.

If at any time, when solving an inequality, multiplication or division by a negative quantity is performed then the inequality symbol reverses.

Let’s look at an example.

Solve the following inequality:

\begin{align*}5-3y \ge 23\end{align*}

First, subtract 5 from both sides of the inequality and simplify.

\begin{align*}\begin{array}{rcl} 5-3y &\ge& 23\\ 5-5-3y &\ge& 23-5\\ -3y &\ge& 18 \end{array}\end{align*}

Next, divide both sides of the inequality by -3 to solve for the variable ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} -3y &\ge& 18\\ \frac{\overset{1}{\cancel{-3}}}{\cancel{-3}}y &\ge& \frac{\overset{-6}{\cancel{18}}}{\cancel{-3}}\\ y &\le& -6 \end{array}\end{align*}

Notice the inequality sign has been reversed because both sides of the equation were divided by -6.

The solution is any value for the variable less than and equal to -6. This solution can be expressed as:

\begin{align*}y \le -6 \ \text{or} \ \{y|y \le -6, y \in R \}\end{align*}

### Examples

#### Example 1

Earlier, you were given a problem about the trip to the rainforest. It is 180 miles one way so the total distance you must travel is 360 miles. The jeep will hold a maximum of 20 gallons of fuel which is a mixture of gasoline and ethanol. You need to figure out how many gallons of gasoline and ethanol you need to buy for the return trip to the rainforest.

Use a system of linear inequalities to calculate the number of gallons of each fuel type is needed.

First, write a system of linear inequalities to model the problem.

\begin{align*}\text{Let }x = \# \text{ gallons of ethanol}\end{align*}

\begin{align*}\begin{array}{rcl} 17x+21y &\ge& 360\\ x+y &\le& 20 \end{array}\end{align*}

Next, solve the system of linear inequalities using the substitution method.

Next, subtract \begin{align*}y\end{align*} from both sides of the second inequality and simplify.

\begin{align*}\begin{array}{rcl} x+y &\le& 20\\ x+y-y &\le& 20-y\\ x &\le& 20-y \end{array}\end{align*}

Next, substitute the expression for the variable ‘\begin{align*}x\end{align*}’ into the first inequality.

\begin{align*}\begin{array}{rcl} 17x+21y &\ge& 360\\ 17(20-y)+21y &\ge& 360 \end{array}\end{align*}

Next, perform the multiplication to clear the parenthesis and simplify.

\begin{align*}\begin{array}{rcl} 17(20-y)+21y &\ge& 360\\ 340-17y+21y &\ge& 360\\ 340+4y &\ge& 360 \end{array}\end{align*}

Next, subtract 340 from both sides of the inequality and simplify.

\begin{align*}\begin{array}{rcl} 340+4y &\ge& 360\\ 340-340+4y &\ge& 360-340\\ 4y &\ge& 20 \end{array}\end{align*}

Next, divide both sides of the inequality by 4 to solve for the variable ‘\begin{align*}y\end{align*}’.

\begin{align*}\begin{array}{rcl} 4y &\ge& 20\\ \overset{1}{\frac{\cancel{4}y}{\cancel{4}}} &\ge& \frac{\overset{5}{\cancel{20}}}{\cancel{4}}\\ y &\ge& 5 \end{array}\end{align*}

Then, substitute the value 5 into the expression for the variable ‘\begin{align*}x\end{align*}’ and simplify.

\begin{align*}\begin{array}{rcl} x &\le& 20-y\\ x &\le& 20-5\\ x &\le& 15 \end{array}\end{align*}

The optimal solution is the ordered pair

which means 15 gallons of ethanol and 5 gallons of gasoline are needed for the return trip to the rainforest.

#### Example 2

Translate the following statement into an inequality and solve it.

The difference between a number and seven is greater than 12.

\begin{align*}x-7>12\end{align*}

First, add 7 to both sides of the inequality to solve for the variable ‘

’.\begin{align*}\begin{array}{rcl} x-7 &>& 12\\ x-7+7 &>& 12+7\\ x &>& 19 \end{array}\end{align*}

Any value for the variable greater than 19 is a solution for the inequality. The solution can be expressed as:

#### Example 3

Translate the following statement into an inequality and solve it.

Two times a number increased by six is less than twenty.

\begin{align*}2x+6<20\end{align*}

First, subtract 6 from both sides of the inequality and simplify.

\begin{align*}\begin{array}{rcl} 2x+6 &<& 20\\ 2x+6-6 &<& 20-6\\ 2x &<& 14 \end{array}\end{align*}

Next, divide both sides of the inequality by 2 to solve for the variable ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 2x &<& 14\\ \overset{1}{\frac{\cancel{2}}{\cancel{2}}} \ x &<& \frac{\overset{7}{\cancel{14}}}{{\cancel{2}}}\\ x &<& 7 \end{array}\end{align*}

Any value for the variable less than 7 is a solution for the inequality. The solution can be expressed as:

\begin{align*}x<7 \ \text{or} \ \{x|x <7, x \in R \}\end{align*}

#### Example 4

Translate the following statement into an inequality and solve it.

Three less than four times a number is greater than and equal to 41.

\begin{align*}4x-3 \ge 41\end{align*}

First, add three to both sides of the inequality and simplify.

\begin{align*}\begin{array}{rcl} 4x-3 &\ge& 41\\ 4x-3+3 &\ge& 41+3\\ 4x &\ge& 44 \end{array}\end{align*}

Next, divide both sides of the inequality by 4 to solve for the variable ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} 4x &\ge& 44\\ \frac{\overset{1}{\cancel{4}}}{\cancel{4}}x &\ge& \frac{\overset{11}{\cancel{44}}}{{\cancel{4}}} \\ x &\ge& 11 \end{array}\end{align*}

Any value for the variable greater than and equal to 11 is a solution for the inequality. The solution can be expressed as:

\begin{align*}x \ge 11 \ \text{or} \ \{x|x \ge 11, x \in R \}\end{align*}

#### Example 5

The difference between 20 and a number is greater than 12.

\begin{align*}20-x > 12\end{align*}

First, subtract 20 from both sides of the inequality and simplify.

\begin{align*}\begin{array}{rcl} 20-x &>& 12\\ 20-20-x &>& 12-20\\ -x &>& -8 \end{array}\end{align*}

Next, divide both sides of the inequality by -1 to solve for the variable ‘\begin{align*}x\end{align*}’.

\begin{align*}\begin{array}{rcl} -x &>& -8\\ \frac{\overset{1}{\cancel{-1}}x}{\cancel{-1}} &>& \frac{\overset{8}{\cancel{-8}}}{\cancel{-1}}\\ x &<& 8 \end{array}\end{align*}

Notice the inequality sign has been reversed because both sides of the equation were divided by -1.

The solution is any value for the variable less than and equal to 8. This solution can be expressed as:

\begin{align*}y<8 \ \text{or} \ \{y|y <8, y \in R \}\end{align*}

### Review

Translate each statement into an inequality.

1. The difference between two numbers is greater than 8.

2. Half of one number is at least 3 times another number.

3. A quarter the sum of two numbers is less than 15.

4. Seven times one number plus 3 less than another is not more than -16.

5. Six times a number is greater than negative thirty.

6. Five times a number and six is less than or equal to 39.

7. Twelve divided by a number is less than seven.

8. Six times a number and two less than another number is less than or equal to -12.

Which ordered pairs make the inequalities true?

9.

10.

11.

12.

13.

14.

15.

16.