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Compound Inequalities

Compound inequality can combine two inequalities to show that expression lies between two fixed values. Simple inequalities are connected by the conditions 'and' or ‘or’. A solution to a compound inequality is a number (chosen from replacement set) which, when substituted for the variable, makes the inequality true.

Compound Inequality is the coupling of two simple inequalities connected by the conditions 'and' or ‘or’.

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Compound Inequality Definition

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Compound inequality is formed by joining two inequalities with a connective word such as "and" or "or". The solution set of a compound inequality with the connective word "and " is the set of all elements common to the solution sets of both inequalities. The solution set of a compound inequality with the connective word "or" is the union of the solution sets of the two inequalities.

Solving Compound Inequalities

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The solution set of a compound inequality with the connective word "and" is the set of all elements common to the solution sets of both inequalities. Solution for the connective word "or" is the set of all elements of both inequalities. The set of all solutions of an inequality is called the solution set of the compound inequality.

Addition - Subtraction Rule:

If the same number or expression is added to or subtracted from both sides of an inequality, the resulting inequality has the same solution (or solutions) as the original.

Multiplication - Division Rule:

If both sides of an inequality are multiplied or divided by the same positive number, the resulting inequality has the same solution as the original.

If both sides of an inequality are multiplied or divided by the same negative number, the resulting inequality has the same solution ( or solutions) as the original if the symbol of the inequality is reversed.


Graphing Compound Inequalities

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Graphing method is the best method to find the solution for the compound of inequalities. Here, we can solve the system of linear inequalities with single variables.

Let us solve 2x < 42 and 4x < 12

Given 2x < 42 and 4x < 12

Step 1:
Solve each inequality

Case 1: 2x < 42

Divide each side by 2

x < 21

Solution set is {x|x < 21}

Case 2: 4x < 12

Divide each side by 4

x < 3

Solution set is {x|x < 3}

Step 2: Since the solution set of a compound inequality with the connective word "and " is the set of all elements common to the solution sets of both inequalities.

So, find the intersection of the solution sets

{x|x < 21} $\cap$ {x|x < 3} = {x|x < 3}

Step 3:
Graph of the given compound inequality:

Compound Inequality

Solving Compound and Absolute Value Inequalities

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Compound inequalities contain two inequalities into a single statement. Solving a compound inequality requires solving each of the inequalities it contains. To solve an absolute inequality of the form |ax + b| < c, solve the equivalent compound inequality -c < ax + c < c. Similarly, we can solve other inequalities also.

Let us solve |4 - 3x| > 1

Step 1: Given |4 - 3x| > 1

Absolute inequality can be written as

4 - 3x < -1 or 4 - 3x > 1

Step 2: Solve each inequality

Case 1: 4 - 3x < -1

4 - 3x - 4 < -1 - 4 (Subtract 4 from each side)

-3x < -5 (Combine like terms)

x > $\frac{5}{3}$ (Divide each side by -5)

Case 2: 4 - 3x > 1

4 - 3x - 4 > 1 - 4 (Subtract 4 from each side)

-3x > -3 (Combine like terms)

x < 1 (Divide each side by -3)

Step 3: Solution of the given inequality is {x | x > $\frac{5}{3}$ or x < 1 }.

Solving Compound Inequalities with Fractions

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To solve fractional inequalities, it is best to clear the inequality of all fractions by multiplying both sides of the equation by the LCM of all the denominators. Then, solve these inequalities as we solve compound inequalities.

Let us solve $\frac{2}{7} \leq \frac{3x - 2}{2} < \frac{2}{3}$. Write the solution set in interval notation.

Given $\frac{2}{7} \leq \frac{3x - 2}{2} < \frac{2}{3}$

Step 1: $\frac{4}{7} \leq 3x - 2 < \frac{4}{3}$ (Multiply each side by 2)

Step 2: 2 + $\frac{2}{7} \leq 3x < \frac{4}{3}$ + 2 ( Add 2 both the sides)

Step 3: $\frac{16}{7} \leq 3x < \frac{10}{3}$ (Combine like terms)

Step 4: $\frac{16}{21} \leq x < \frac{10}{9}$ (Divide each side by 3)

Step 5: 0.76 $ \leq$ x < 1.11

Step 6: The solution of the given inequality is [0.76, 1.11).

Compound Inequalities Examples

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Given below are the examples based on solving compound Inequalities.

Solved Examples

Question 1: Solve inequalities for 2x + 4 < 16 and 5x - 7 > - 12.
Solution:

Solve each Compound Inequality separately. Since the combination expression is "and", this indicates that the overlap or intersection is the desired result.

2x + 4 < 16 and 5x - 7 > - 12

2x < 12 and 5x > -5

x < 6 and x > -1

x < 6 indicates all the numbers to the left of 6, and x > -1 indicates all the numbers to the right of -1. The intersection of all these numbers between -1 and 6. The solution set is

{x|x > -1 and x < 6}

The solution set could be expressed as {x |-1 < x < 6}.



Question 2: Find the solution set of 2 $\leq$ 3 ( x - 2) + 5 < 8, x $\in$ W. Also represent its solution on the number line.
Solution:

Given 2 $\leq$ 3 ( x - 2) + 5 < 8

Step 1: 2 $\leq$ 3 x - 6 + 5 < 8 (Using distributive property)

Step 2: 2 $\leq$ 3x - 1 < 8 (Combine like terms)

Step 3: 2 + 1 $\leq$ 3x - 1 + 1 < 8 + 1 (Add 1 to both the sides)

Step 4: 3 $\leq$ 3x < 9 (Combine like terms)

Step 5: 1 $\leq$ x < 3 (Divide each side by 3)

Step 6: Since x $\in$ W, the solution of the given inequality is x $\geq$ 0.

Solution of the inequality represented on number line:

Compound Inequalities



Compound Inequalities Practice Problems

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Given below are the practice problems on compound Inequalities.

Practice Problems

Question 1: Solve the compound inequality (2x + 1) / 3 $\geq$ (3x - 2) / 5, x $\in$ R. Graph the solution set on the number line.
Question 2: Solve the inequality 3 - 2x $\geq$ x - 12, given that x $\in$ N.
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