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A compound inequality calculator displays the inequality equation with number line representation when given a compound inequality. In STUDYQUERIES’s online compound inequality calculator tool, the calculation is faster, and the inequality equation is displayed in just a few seconds.

**How to Use the Compound Inequality Calculator?: Step By Step Guide**

The procedure to use the compound inequality calculator is as follows:

**Step 1:**Enter two inequality equations in the respective fields**Step 2:**Click the “Solve” button to get the inequality**Step 3:**The inequality equation with the number line will be displayed in a new window

Compound Inequality Calculator

**What Is Compound Inequality And Compound Inequality Calculator?**

The compound inequality is made up of two inequality statements connected either by the word “or” or by the word “and.” “And” indicates that both statements are true at the same time. In other words, the solution sets for the individual statements overlap or intersect. “Or” indicates that if either statement is true, the entire compound sentence is true. This solution set is the combination or union of the solution sets for the individual statements.

The inequality that uses the word “and” is called a conjunction. While “and” and “or” are parts of speech known as conjunctions, the mathematical conjunction has a different meaning than the grammatical one. In order to illustrate this point, the conjunction “or”, when used in a compound inequality, creates a disjunction. Remember that “con” means “with another” and “dis” means “one OR the other.”

**Solving Methods Of Compound Inequalities**

We will deal with two general types of compound inequalities when solving them.

- For the first problem, we must solve two linear inequalities connected by the word “and”. The conjunction “and” is also known as a conjunction. The solution of an “and” compound inequality is the set of all values of x that satisfy both the inequalities. In other words, you need a solution set that works with both inequalities. Another way of saying it is that the solution set of the “and” compound inequality is the intersection, represented by the symbol \(\Large{\Bbb \cap}\), of the two inequalities.
- In the second case, you must solve two linear inequalities that are connected by the word “or”. If there is an “or” compound inequality, the solution is the set of all x that satisfies one or both inequalities, or sometimes both at the same time. In other words, you want a solution that works on at least one inequality. It can also be said that the solution set of an “or” compound inequality is the union, represented by the symbol \(\Large{\Bbb \cup}\), of the two inequalities.

Both graphs on the number line as well as interval notation can be used to express the solutions of compound inequalities.

Prior to expressing the solution of the compound inequality in interval notation, you should graph the solutions of the two inequalities on the number line. With an understanding of how the two inequalities behave on the number line, it is much easier to write the interval notation.

Additionally, we will look at some examples where compound inequality has no solution or has an infinite solution.

We will discuss an “and” compound inequality case somewhere in our examples, which can be condensed into a single inequality with three parts: the left side, the middle part, and the right side. An example would be \(- 1 \le x \le 3\) which is derived from \(-1 \le x\) and \(x \le 3\). By writing it in this form, we can solve the compound inequality more quickly.

**The “AND” Compound Inequalities**

Solve the compound “and” inequality by solving each of the two inequalities separately, then look at their solutions together. For the “and” case, we want to find all the numbers or values that can make both inequalities true.

**Solve the compound inequality **\(x – 1 \gt 1\)** and **\(27 \ge 2x – 1\)**. Graph the solutions on the number line. Then, write your solutions in interval notation.**

**Step 1: Solve each inequality.**

**First inequality:** \(x-1 \gt 1\)

Add 1 to each side of the inequality.

$$x – 1+1 \gt 1+1$$

$$\Bbb x \gt 2$$

**Second inequality:** \(27\ge 2x-1\)

Divide both sides of the inequality by 2 after adding both sides by 1. Make sure the variable is on the left. In this case, the variable x will move from left to right when you swap locations. In order to maintain the meaning of the inequality symbol, it should remain in the same relative orientation.

It can be seen that the “mouth” of the inequality symbol is opening towards the number 14. When you swap, the mouth of the inequality must still point towards 14.

$$27 \ge 2x – 1$$

$$27 + 1 \ge 2x – 1 + 1$$

$$28 \ge 2x$$

$$\frac{28}{2} \ge \frac{2x}{2}$$

$$14 \ge x$$

$$x \le 14$$

The solutions are given by \(\Bbb {x} \gt 2\) and \(\Bbb {x} \le 14\).

**Step 2: Graph the solutions on the number line.**

For \(x \gt 2\), point 2 is not included as part of the solutions since \(x \gt 2\) means all numbers greater than 2. Further, it does not have any conditions of equality, so we must exclude the number 2. An open circle will be placed over 2 to indicate that it is not a solution. We draw an arrow to the right of 2 because all the solutions are greater than 2.

For \(x \le 14\), we read it as “x is less than or equal to 14“. There is an equality condition, therefore the number 14 is part of the solution, so we’ll close the circle around it. We will draw an arrow to the left of 14 to indicate that all numbers to the left of it are also solutions.

The final solutions will be the intersection or overlap of the two inequalities: \(x \gt 2\) and \(x \le 14\). All the numbers between 2 and 14 are part of the final solutions to the “and” compound inequality. We add the number 14 to the solution set because they intersect at the number 14. However, they do not intersect at point 2, so we drop it as part of the solution. We have just determined the solution set for the given compound inequality.

On the number line, all values of x greater than 2 but less than or equal to 14 are shown as a compound inequality

**Step 3: Write the solutions in interval notation.**

All numbers between 2 and 14 are included in the solution. Moreover, the number 2 is excluded since it has an open circle, while the number 14 is included since it has a closed circle. As of now, we use a rounded bracket or parenthesis if it is excluded (2 is excluded), and a square bracket if it is included (14 is included).

\(\left( {2,14} \right]\)

This is read as “all the numbers greater than 2 but less than or equal to 14.”

$$\mathit{\color{red}{This\ type\ of\ interval\ is\ also\ known\ as\ half-closed\ or}}$$

$$\mathit{\color{red}{half-open\ interval\ because\ one\ of\ the\ two\ endpoints\ is\ included\ but\ the\ other\ is\ not.}}$$

** Solve the compound inequality **\(2+3x\gt -10\)

**\(2\left( {x – 1} \right) \lt x + 4\)**

*and*

*. Graph the solution set on the number line. Then, write the solution set in the interval notation.***Step 1: Solve each inequality.**

**First inequality:** \(2 + 3x \gt – 10\)

Inequalities on both sides are subtracted by 2. Divide both sides by 3.

$$2 + 3x \gt – 10$$

$$2 – 2 + 3x \gt – 10 – 2$$

$$3x \gt – 12$$

$$\frac{3x}{3} \gt \frac{-12}{3}$$

$$x\gt −4$$

**Second Inequality:** \(2\left( {x – 1} \right) \lt x + 4\)

Distribute the 2 to the binomial inside the parenthesis. Add 2 on both sides by of the inequality. Then subtract both sides by x.

$$2\left( {x – 1} \right) \lt x + 4$$

$$2x – 2 \lt x + 4$$

$$2x – 2 + 2 \lt x + 4 + 2$$

$$2x \lt x + 6$$

$$2x – x \lt x – x + 6$$

$$x \lt 6$$

The solutions are given by \(x\gt −4\) and \(x \lt 6\).

**Step 2: Graph the solution set on the number line.**

Strict inequality is a type of inequality that is either absolutely greater than a number, \(x\gt a\), or absolutely lesser than a number, \(x\lt a\). You will notice that strict inequality does not contain an equality component.

On the other hand, the inequality symbol \(x \ge a\) which is read as “x is greater than or equal to a, and the inequality symbol \(x \le a\) which is read as “x is less than or equal to a” are both non-strict inequalities because they have the equality conditions.

The inequality \(x\gt −4\) is a strict inequality therefore we will put an open circle over -4 as it is not part of the solutions, and draw an arrow to the right. Similarly, \(x\lt 6\) is a strict inequality thus we will put an open circle over 6, and draw an arrow to the left.

The final solution set will be the intersection of \(x\gt −4\) and \(x\lt 6\) which are all the numbers between −4 and 6 but excluding the endpoints -4 and 6.

**Step 3: Write the solutions in interval notation.**

On both sides, we will use rounded brackets or parentheses to indicate that both endpoints are excluded from the solution set.

$$\left( {-4,6} \right)$$

It is read as “all the numbers greater than -4 but less than 6.

\(\mathit{\color{red}{This\ type\ of\ interval\ is\ also\ known\ as\ an\ open\ interval\ because\ the\ two\ endpoints\ are}}\)

\(\mathit{\color{red}{excluded\ in\ the\ solution\ set.\ That\ is,\ they\ are\ NOT\ part\ of\ the\ solutions.}}\)

* Solve the compound inequality *\(5 – 3\left( {x – 2} \right) \le x – \left( { – 2x + 13} \right)\)

*\(5 – \left( {x + 1} \right) \le 2\left( {7 – x} \right) + 1\)*

**and**

**. Graph the solution set then write its solutions in the interval notation.****Step 1: Solve each inequality.**

**First inequality:** \(5 – 3\left( {x – 2} \right) \le x – \left( { – 2x + 13} \right)\)

Using the Distributive Property of Multiplication over Addition, remove the parenthesis from each side of the inequality. Put 5 and 6 on one side of the inequality. Subtract 1 from both sides. Subtract both sides by 3x. Divide both sides by -6. As we divide each side by a negative number, we will switch the direction of the inequality. Hence, “from less than or equal to” to “greater than or equal to”.

$$5 – 3\left( {x – 2} \right) \le x – \left( { – 2x + 13} \right)$$

$$5 – 3x + 6 \le x + 2x – 13$$

$$- 3x + 11 \le 3x – 13$$

$$- 3x + 11 – 11 \le 3x – 13 – 11$$

$$- 3x \le 3x – 24$$

$$- 3x -3x \le 3x – 3x+24$$

$$- 6x \le-24$$

$$\frac{- 6x}{-6} \le \frac{-24}{-6}$$

$$x \ge 4$$

**Second Inequality:** \(5 – \left( {x + 1} \right) \le 2\left( {7 – x} \right) + 1\)

Using the Distributive Property of Multiplication over Addition, remove the parentheses. Subtract 5 by 1 on the left side. Add 4 to both sides of the inequality. Add 2x to both sides to finish it off.

$$5 – \left( {x + 1} \right) \le 2\left( {7 – x} \right) + 1$$

$$5 – x – 1 \le 14 – 2x + 1$$

$$4 – x \le 15 – 2x$$

$$4 – 4 – x \le 15 – 4 – 2x$$

$$- x \le 11 – 2x$$

$$- x + 2x \le 11 – 2x + 2x$$

$$x \le 11$$

The solutions are given by \(x \ge 4\) and \(x \le 11\).

**Step 2: Graph the solution set on the number line.**

For \(x \ge 4\), we will shade the circle above 4 to show that it is included in the solutions because the inequality has a condition of equality, that is, “greater than or equal to”. The arrow points to the right of 4 since it has a component of “greater than”.

For \(x \le 11\), we will also shade the circle above 11 to indicate that is it is part of the solution set since the inequality has a condition of equality, that is, “less than or equal to”. Due to the case of less than, the arrow points to the left of 11.

In the final solution set, we find all the points where the two inequalities intersect. It is obvious that they intersect between 4 and 11. They also overlap at the endpoints. Accordingly, the final solution set contains all the points between endpoints 4 and 11 and includes the endpoints.

**Step 3: Write the solution set in interval notation.**

Square brackets will be used on both sides to indicate that both endpoints are included in the solution set.

$$\left[ {4,11} \right]$$

It is read as “all numbers greater than or equal to 4 but less than or equal to 11.

\(\mathit{\color{red}{This\ type\ of\ interval\ is\ also\ known\ as\ a\ closed\ interval\ because\ the\ two\ endpoints\ are}}\)

\(\mathit{\color{red}{included\ in\ the\ solution\ set.\ That\ is,\ they\ are\ part\ of\ the\ solutions.}}\)

** Solve the compound inequality **\(3x – 2\left( {1 – x} \right) \lt x – 6\)

**\(10 – x \lt x + 2\)**

*and*

*. Graph the solution set then write its solutions in the interval notation.***Step 1: Solve each inequality.**

**First inequality:** \(3x – 2\left( {1 – x} \right) \lt x – 6\)

Remove the parenthesis by distributing the -2 into the binomial 1-x on the left side of the inequality. Also, add 3x and 2x to the left side. Divide by 2 into both sides. Subtract x from both sides. Divide by 4 into both sides.

$$3x – 2\left( {1 – x} \right) \lt x – 6$$

$$3x – 2 + 2x \lt x – 6$$

$$5x – 2 \lt x – 6$$

$$5x – 2 + 2 \lt x – 6 + 2$$

$$5x \lt x – 4$$

$$5x – x \lt x – x – 4$$

$$4x \lt – 4$$

$$\frac{4x}{4}\lt \frac{-4}{4}$$

$$x \lt -1$$

**Second inequality:** \(10 – x \lt x + 2\)

Subtract both sides of the inequality by 10. Then, subtract it also on both sides by x. Finally, divide each side by -2. Since we are dividing by a negative number, we must flip or switch the direction of the inequality symbol. In this case, from less than to greater than.

$$10 – x \lt x + 2$$

$$10 – 10 – x \lt x + 2 – 10$$

$$- x \lt x – 8$$

$$- x – x \lt x – x – 8$$

$$- 2x \lt – 8$$

$$\frac{-2x}{- 2}\gt \frac{-8}{-2}$$

$$x \gt 4$$

Observe that the solutions of the two inequalities \(x \lt- 1\) and \(x \gt 4\) do not intersect, and thus the compound inequality has no solution.

Their graphs on the number line make it more obvious that they do not overlap.

**The “OR” Compound Inequalities**

Solve the compound “or” inequality by solving each of the two inequalities separately. For the “or” case, we want to find all the numbers that can make at least one of the two inequalities to be true.

** Solve the compound inequality **\(2x – 5 \gt 3x + 2\)

**\(x – 1 \lt 2x – 5\)**

*or*

*. Graph the solutions on the number line. Then, write your solutions in interval notation.***Step 1:** Solve each inequality.

Add 5 to both sides of the inequality. Then subtract 3x on both sides. Finally, divide -1 into both sides. Don’t forget to flip the inequality symbol because we divided a number by a negative number.

**First inequality:** \(2x – 5 > 3x + 2\)

$$2x – 5 + 5 > 3x + 2 + 5$$

$$2x > 3x + 7$$

$$2x – 3x > 3x – 3x + 7$$

$$-x > 7$$

$$\frac{-x}{- 1} < \frac{7}{-1}$$

$$x < – 7$$

**Second inequality:** \(x – 1 < 2x – 5\)

Add both sides by 1. Then subtract by 2x to both sides. Divide both sides of the inequality by -1 thus switching the direction of the inequality symbol.

$$x – 1 < 2x – 5$$

$$x – 1 + 1 < 2x – 5 + 1$$

$$x < 2x – 4$$

$$x – 2x < 2x – 2x – 4$$

$$- x < – 4$$

$$\frac{-x}{-1} > \frac{4}{-1}$$

$$x > 4$$

The solutions are given by \(x < – 7\) or \(x > 4\).

**Step 2:** Graph the solution set on the number line.

**Step 3:** Write the solutions in interval notation.

\(\left( { – \infty ,7} \right) \cup \left( {4,\infty } \right)\)

It is as read as “all numbers less than negative 7 or all numbers greater than 4.

** Solve the compound inequality **\(2\left( {x + 1} \right) \le x – 2\)

**\(3\left( {x – 1} \right) \le 4x – 3\)**

*or*

*. Graph the solutions on the number line. Then, write your solutions in interval notation.***Step 1:** Solve each inequality.

**First inequality:** \(2\left( {x + 1} \right) \le x – 2\)

Distribute 2 into the quantity (x+1). Subtract 2 on both sides of the inequality. Finally, subtract xx on both sides to get to the final solution.

**Second inequality:** \(3\left( {x – 1} \right) \le 4x – 3\)

Distribute 3 into the quantity \((x-1)\). Next, add 3 to both sides of the inequality. Then, subtract sides by 4x. Finally divide both sides by -1. Please don’t forget to switch the direction of the inequality from “less than or equal to” to “greater or equal to”.

The solutions are given by \(x \le – 4\) or \(x \ge 0\).

**Step 2:** Graph the solution set on the number line.

**Step 3:** Write the solutions in interval notation.

\(\left({ – \infty , – 4} \right) \cup \left( {0,\infty } \right)\)

It is read as “all numbers less than or equal to -4 or all numbers greater than or equal to 0.

**FAQs**

**What is a compound inequality give some examples?**

Compound inequalities are the derived form of inequalities, which are very useful in mathematics whenever dealing with a range of possible values. For example, after solving a particular linear inequality, you get two solutions, x>3 and x<12. You can read it as “3 is less than x, which is less than 12.

**What is a compound inequality and how is it solved?**

A compound inequality contains at least two inequalities that are separated by either “and” or “or”. The graph of a compound inequality with an “and” represents the intersection of the graph of the inequalities. A number is a solution to the compound inequality if the number is a solution to both inequalities.

**Are compound inequalities AND or OR?**

A compound inequality is a sentence with two inequality statements joined either by the word “or” or by the word “and.” “And” indicates that both statements of the compound sentence are true at the same time. “Or” indicates that, as long as either statement is true, the entire compound sentence is true.

**How do you tell if it is an AND or OR inequality?**

If it is a conjunction that uses the word and, the solution must work in both inequalities and the solution is in the overlap region of the graph. If it is a disjunction that uses the word or, the solution must work in either one of the equations.

**What inequality has no solution?**

Here are the steps to follow when solving absolute value inequalities: Isolate the absolute value expression on the left side of the inequality. If the number on the other side of the inequality sign is negative, your equation either has no solution or all real numbers as solutions.