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 direction.} & \frac{-4x}{-4} & > & \frac{-6}{-4}\\ \text{Simplify.} & x & > & \frac{3}{2}\\ \end{array}[/latex]

Exercise 1


Example 3

A company spends $1,200 per day on overhead and labor, and each item they produce costs $5 for materials. If they sell the items for $10 each, how many items will they need to sell each day for their profits to be positive?

While we could solve this problem using equations, it also lends itself to inequalities, since we want the profit to be positive: [latex]P > 0[/latex].

Costs: [latex]C(q) = 1200 + 5q[/latex]                    

Revenue: [latex]R(q) = 10q[/latex]

Profit: [latex]p(q) = 10q - (1200 + 5q) = 5q - 1200[/latex]


Solving [latex]P(q) > 0[/latex]:

[latex]\begin{array}{rcl} 5q-1200 & > & 0 \\ 5q & > & 1200\\ q & > & 240\\ \end{array}[/latex]

The company will need to sell at least 240 items a day to make a profit.

Compound Inequalities

Compound inequalities are inequalities that consist of more than one part. The most common type is called a tripartite inequality. The basic version looks like this: [latex]-1 \lt 3x+5 \lt 14[/latex].

When we write these, it is important that both inequalities point in the same direction and that the “outside” inequality is also true—in this case, [latex]-1 \lt 14[/latex] is true, so this is valid. Expressions like [latex]10 \lt x \lt 2[/latex] and [latex]1 \lt x > 5[/latex] are not valid notation.

The most universal way to solve a tripartite inequality is to:

  1. Break it into two separate inequalities.
  2. Solve each inequality separately.
  3. Combine the solutions if possible.

Example 4

Solve [latex]-1 \lt -3x+5 \lt 14[/latex].

First we separate this into two inequalities: [latex]-1 \lt -3x+5[/latex] and [latex]-3x+5 \lt 14[/latex].

Now we solve each:

[latex]\begin{array}{rclcrcl} -1 & \lt & -3x+5 & & -3x+5 & \lt & 14\\ -6 & \lt & -3x & & -3x & \lt & 9\\ 2 & > & x & & x & > & -3\\ \end{array}[/latex]

Now we can combine these solution sets. The numbers where both [latex]2 > x[/latex] and [latex]x > -3[/latex] are true is the set: [latex]2 > x > -3[/latex].

While this solution is valid and correct, it is more common to write the solution to tripartite inequalities with the smaller number on the left. We could rewrite the solution as [latex]-3 \lt x \lt 2[/latex].

This also has the advantage of corresponding better with the answer in interval notation: [latex](-3, 2)[/latex].

With this particular inequality, it would also be possible to skip the step of breaking it apart and instead just subtract 5 from all three “parts” of the inequality. This works for simple problems like this but may fail if the inequality has variables in more than one “part” of the inequality.

Exercise 2

Solve [latex]4 \leq 2x+6 \lt 16[/latex].


[latex]-1 \leq x \lt 5[/latex] or [latex][-1,5)[/latex]


Solve [latex]-5 \lt -2x-4 \lt 0[/latex].


[latex]-2 \lt x \lt \frac{1}{2}[/latex] or [latex](-2,\frac{1}{2})[/latex]



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