Chapter 3 Linear Programming
3.1 Inequalities in One Variable
Learning Objectives
By the end of this section, you will be able to:
- Solve linear inequalities
- Solve compound inequalities
In this section, we will explore how to solve linear and absolute value inequalities in one variable. The process is very similar to solve equations, but instead of the solution being a single value, the solution will be an inequality.
Notice that if an inequality is true, like [latex]2 \lt 5[/latex], then these operations result in a true statement as well, just like with equations:[latex]\begin{array}{cccc} \text{Adding a number} & \text{Subtracting a number} & \text{Multiplying a positive} & \text{Dividing a positive}\\ \text{to both sides:} & \text{from both sides:} & \text{number on both sides:} & \text{number on both sides:}\\ 2+4 \lt 5+4 & 2-3 \lt 5-3 & 2(3) \lt 5(3) & \frac{2}{2} \lt \frac{5}{2}\\ 6 \lt 9 & -1 \lt 2 & 6 \lt 15 & 1 \lt 2.5\\ \text{True} & \text{True} & \text{True} & \text{True}\\ \end{array}[/latex]
We can use these operations just like when solving equations.
Example 1
Solve [latex]3x+7 \geq 1[/latex]
[latex]\begin{array}{lrcl} & 3x+7 & \geq & 1\\ \text{Subtract 7 from both sides.} & 3x+7-7 & \geq & 1-7 \\ \text{Simplify.} & 3x & \geq & -6\\ \text{Divide both sides by 3.} & \frac{3x}{3} & \geq & \frac{-6}{3}\\ \text{Simplify.} & x & \geq & -2 \end{array}[/latex]
This inequality represents the solution set. It tells us that all numbers greater than or equal to -2 will satisfy the original inequality. We could also write this solution in interval notation, as [latex][-2,\infty)[/latex].
To understand what is happening, we could also consider the problem graphically. If we were to graph the equation [latex]y=3x+7[/latex], then solving [latex]3x+7 \geq 1[/latex] would correspond with asking, “For what values of [latex]x[/latex] is [latex]y \geq 1[/latex]?”
Notice that the part of the graph where this is true corresponds to where [latex]x \geq -2[/latex].
While most operations in solving inequalities are the same as in solving equations, we run into a problem when multiplying or dividing both sides by a negative number. Notice, for example:
[latex]\begin{array}{rcl} 2(-3) & \lt & 5(-3)\\ -6 & \lt & -15\\ &\text{False}\\ \end{array}\\[/latex]
To account for this, when multiplying or dividing by a negative number, we must reverse the sign of the inequality.
Rules for Solving Linear Inequalities
- You may add or subtract a positive or negative number to both sides of the inequality.
- You can multiply or divide both sides of the inequality by a positive number.
- You can multiply or divide both sides of the inequality by a negative number, but you must reverse the direction of the inequality.
Example 2
Solve [latex]12-4x \lt 6[/latex]
[latex]\begin{array}{lrcl} & 12-4x & \lt & 6\\ \text{Subtract 12 from both sides.} & 12-4x-12 & \lt & 6-12 \\ \text{Simplify.} & -4x & \lt & -6\\ \text{Divide both sides by -4 and reverse }\\ \text{the inequality