# Solving Compound Inequalities

When solving compound inequalities, we are going to deal with two general cases or types.

- The first case involves solving two linear inequalities joined by the word “and”. The word “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 of the two inequalities. In other words, you want a solution set that works with both inequalities. The other way of saying it is that the solution to the “and” compound inequality is the intersection, represented by the symbol \Large{\color{red} \cap}, of the solutions of the two individual inequalities.

- As for the second case, it involves solving two linear inequalities joined by the word “or”. The solution of an “or” compound inequality is the set of all x that satisfy either of the two inequalities or at times satisfy the two at the same time. In other words, you want a solution that works on at least one inequality. The other way of saying it is that the solution to the “or” compound inequality is the union, represented by the symbol \Large{\color{red} \cup }, of the solutions of the two individual inequalities.

For both cases, the solutions of compound inequalities can be expressed as graphs on the number line and also as interval notations.

I suggest that you first graph the solutions of the two inequalities on the number line before writing the solution of the compound inequality in the interval notation. By having a visual representation on how the two inequalities behave on the number line, it is much easier writing its corresponding interval notation.

We will also go over some examples where the compound inequality has no solution or infinite solution.

Finally, I will show you that sometimes two inequalities of an “and” case can be abbreviated into a single inequality with three parts: left side, center part, and right side. This is also known as** double inequality**. 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, it can allow us to solve the compound inequality much quicker.

## The “AND” Compound Inequality

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

**Example 1:** Solve the compound inequality x - 1 > 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 > 1

Add 1 to both sides of the inequality.

x - 1 > 1

x - 1+1 > 1+1

\color{red}x > 2

- Second inequality: 27 \ge 2x - 1

Add both sides of the inequality by 1 then divide by 2. Finally, make sure that the variable is on the left side. When you swap location, in this case, the variable **x** will move from right to left. The relative orientation of the inequality symbol should remain the same to keep the meaning unchanged. One way to think about it is that the “mouth” of the inequality symbol is opening towards the number 14. So when you swap, the “mouth” of the inequality must still be pointing towards 14.

27 \ge 2x - 1

27 + 1 \ge 2x - 1 + 1

8 \ge 2x

{\Large{{{28} \over 2}}} \ge {\Large{{{2x} \over 2}}}

14 \ge x

\color{red}x \le 14

The solutions are given by \color{red}x > 2 and \color{red}x \le 14.

**STEP 2.** Graph the solutions on the number line.

For \color{red}x > 2, the point 2 is not included as part of the solutions since x > 2 means all numbers greater than 2. In addition, it does not have any conditions of equality that’s why we must exclude the number 2. So we will put an open circle over 2 to indicate that it is not a solution. The solutions are all numbers greater than 2, thus we draw an arrow to the right of 2.

For \color{red}x \le 14, we read it as “x is less than or equal to 14“. Notice there is a condition of equality, therefore the number 14 is part of the solution so we will put a closed circle over it. All numbers to the left of 14 are also solutions so we will draw an arrow pointing to the left of it.

The final solutions will be the intersection or overlap of the two inequalities: \color{red}x > 2 *and* \color{red}x \le 14. Notice that all the numbers between 2 and 14 intersect so they are part of the final solutions of the “and” compound inequality. They also intersect at the number 14 so we add it in the solution set. However, they do not intersect at point 2, thus we drop it off as part of the solutions. We have just figured out the complete solution set of the given compound inequality.

**STEP 3.** Write the solutions in interval notation.

Observe that all numbers between 2 and 14 are part of the solutions. In addition, the number 2 is excluded because it is with an **open circle** while 14 is included because it covered with a **closed circle**. Now, we use a rounded bracket or parenthesis if it is excluded (2 is excluded), and use a square bracket if included (14 is included).

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

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

**Remember:** This type of interval is also known as **half-closed** or half-open interval because one of the two endpoints is included but the other is not.

**Example 2:** Solve the compound inequality 2 + 3x > - 10 *and* 2\left( {x - 1} \right) < x + 4. 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 > - 10

Subtract both sides of the inequality by 2. Then divide both sides by 3.

2 + 3x > - 10

2 - 2 + 3x > - 10 - 2

3x > - 12

{\Large{{{3x} \over 3}}} > {\Large{{{ - 12} \over 3}}}

\color{red}x > - \,4

- Second Inequality: 2\left( {x - 1} \right) < 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) < x + 4

2x - 2 < x + 4

2x - 2 + 2 < x + 4 + 2

2x < x + 6

2x - x < x - x + 6

\color{red}x < 6

The solutions are given by \color{red}x > - \,4 and \color{red}x < 6.

**STEP 2.** Graph the solution set on the number line.

A **strict inequality** is a type of inequality that is either absolutely greater than a number, x>a, or absolutely less than a number, x<a. Notice that strict inequality does not contain any 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 \color{red}x > - \,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, \color{red}x < 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 \color{red}x > - \,4 and \color{red}x < 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.

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

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

**Remember:** This type of interval is also known as an **open interval** because the two endpoints are excluded in the solution set. That is, they are NOT part of the solutions.

**Example 3:** Solve the compound inequality 5 - 3\left( {x - 2} \right) \le x - \left( { - 2x + 13} \right) and 5 - \left( {x + 1} \right) \le 2\left( {7 - x} \right) + 1. 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)

Get rid of the parenthesis on each side of the inequality using the Distributive Property of Multiplication over Addition. Add 5 and 6 on the left. Subtract both sides by 11. Subtract both sides by 3x. To solve x, divide both sides by -6. Since we divide each side by a negative number, we will switch the direction of the inequality. That is, “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

{\Large{{{ - 6x} \over { - 6}}}} \le {\Large{{{ - 24} \over { - 6}}}}

x \ge 4

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

Eliminate the parentheses using the Distributive Property of Multiplication over Addition. Subtract 5 by 1 on the left side. Subtract 4 to both sides of the inequality. Then 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 \color{red}x \ge 4 and \color{red}x \le 11 .

**This is a work in progress. Thank you for your patience.**