Explainer: Compound Linear Inequalities

In this explainer, we will learn how to solve compound linear inequalities by applying inverse operations.

Inequalities are useful for giving us information about how two quantities are related when they are not necessarily equal. Often, a single inequality does not give us enough information regarding the relationship between two quantities; in such cases, we might use compound inequalities. In general, compound inequalities give us two or more conditions that hold either separately or simultaneously.

Compound Inequality

A compound inequality can be presented in two forms. It can be two inequalities connected by an β€œand,” which are often presented together and can be separated into two separate inequalities, an example being the compound inequality 11<7π‘₯+4≀32.

This is an amalgamation of the inequalities 7π‘₯+4>11 and 7π‘₯+4≀32.

Alternatively, compound inequalities can be connected by an β€œor”; for example, 5π‘₯<157π‘₯+3β‰₯52.or

Here, the variable π‘₯ needs to satisfy either the first or the second inequality.

To solve a compound inequality, we often solve each inequality separately and then combine the solutions. We will begin by looking at the inequalities introduced above.

Example 1: Solving Compound Linear Inequalities

Find all the values of π‘₯ that satisfy 11<7π‘₯+4≀32. Give your answer in interval form.

Answer

We can solve this inequality by solving each of the inequalities separately in turn. First, we consider 7π‘₯+4>11.

By subtracting four from each side, we get 7π‘₯>7.

We can now divide each side by 7 to find that π‘₯>1.

Now, we need to solve 7π‘₯+4≀32.

We subtract four from each side to get 7π‘₯≀28 and then divide each side by seven to get π‘₯≀4.

If we combine the two inequalities, we can see that π‘₯ is greater than one but less than or equal to four: 1<π‘₯≀4.

We could represent this solution as an interval, (1,4], or on a number line:

Example 2: Solving Compound Inequalities

Find all the values of π‘₯ that satisfy 5π‘₯<15 or 7π‘₯+3β‰₯52.

Answer

We can solve this inequality by solving each of the inequalities separately in turn. We look first at 5π‘₯<15.

To solve the first inequality, we need to divide each side by five to get π‘₯<3.

To solve the second inequality, we first need to subtract three from each side to get 7π‘₯β‰₯49, and then we divide each side by seven: π‘₯β‰₯7.

So, we know that π‘₯ is less than three or greater than or equal to seven. These inequalities can not be combined, so we present our solutions together: π‘₯<3π‘₯β‰₯7.or

We could also present our solution on a number line:

In the case of inequalities that hold simultaneously, it is sometimes possible to solve them together rather than solve each inequality independently. In the next example, we will demonstrate this method.

Example 3: Solving Compound Inequalities

Find all the values of π‘₯ that satisfy the inequality 15≀3π‘₯+3<21.

Answer

We could separate this into two inequalities and solve each independently. However, it is also possible to solve them together as we will demonstrate here. We start by subtracting three from all parts of the inequality to get 12≀3π‘₯<18.

We can then divide each of the three parts by three, which yields 4≀π‘₯<6.

With the approach we used in the previous example, we need to be particularly careful with negative coefficients. This is because, as you will recall, when we divide or multiply an inequality with a negative number, we need to reverse the inequality sign.

Let us have a look at a more formal example where this is the case.

Example 4: Solving Compound Linear Inequalities

Find all values of π‘₯ that satisfy βˆ’18β‰€βˆ’π‘₯+4β‰€βˆ’1. Write your answer as an interval.

Answer

There are two approaches for solving this compound inequality. The first is to separate it into two inequalities and then solve them individually. Consider first βˆ’π‘₯+4β‰₯βˆ’18.

We add π‘₯ and 18 to each side to get 22β‰₯π‘₯.

Now, consider βˆ’π‘₯+4β‰€βˆ’1.

Here, we add π‘₯ and one to each side to get 5≀π‘₯.

Combining these to inequalities, we see that 5≀π‘₯≀22.

We can write this in interval notation as follows: [5,22].

Alternatively, we could solve this more directly but we would need to be careful when dividing though by βˆ’1. We will demonstrate this now. Consider again the inequality βˆ’18β‰€βˆ’π‘₯+4β‰€βˆ’1.

We subtract four from all three parts to get βˆ’22β‰€βˆ’π‘₯β‰€βˆ’5.

Then, we divide all three parts through by βˆ’1. In doing this, however, we need to reverse the inequalities to get 5≀π‘₯≀22.

This can then be written as an interval as follows: [5,22].

Let us now look at an example where we have variables on each side of the compound inequality.

Example 5: Solving Compound Linear Inequalities with a Variable on Each Side

Find all values of π‘₯ that satisfy 4+π‘₯<2π‘₯βˆ’5<7+π‘₯. Write your answer as an interval.

Answer

Our first step is to subtract π‘₯ from each part of the inequality to get 4<π‘₯βˆ’5<7.

We then need to add five throughout the inequality, giving us 9<π‘₯<12.

If we then write this as an interval, we get (9,12).

Before finishing, let us now look at a pair of compound inequalities that contain an intersection of values. When this is the case, a decision needs to be made as to how the final solution is presented.

Example 6: Solving Compound Inequalities with Overlapping Solutions

Find all the values of π‘₯ that satisfy the compound inequality 7π‘₯βˆ’5>βˆ’126π‘₯+3β‰₯15.or

Answer

To solve the inequality on the left first, we add five to each side: 7π‘₯>βˆ’7.

Then, we divide both sides by seven to find that π‘₯>βˆ’1.

To solve the second inequality, we first subtract three from each side to get 6π‘₯β‰₯12.

Then, we divide each side by six to find that π‘₯β‰₯2.

We can then present the two solutions on a number line:

As the compound inequalities are connected by an β€œor,” our final solution needs to satisfy at least one of the inequalities. So, our solution is π‘₯>βˆ’1.

Were the inequalities connected by an β€œand,” our final solution would need to satisfy both inequalities, in which case the solution would have been π‘₯β‰₯2.

Key Points

  • Compound inequalities give us two or more conditions that hold either separately or simultaneously.
  • Generally, we solve compound inequalities by solving each inequality separately and then combining the solutions.
  • Compound inequalities can have no solution, a single unique solution, or infinitely many solutions.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.