Introduction.

The compound inequality is used in many areas such as engineering appliacations,quadratic equations and

trignometric graphs to find the range and domain of these functions.

The compound inequality is used in many areas such as engineering appliacations,quadratic equations and

trignometric graphs to find the range and domain of these functions.

An inequality is a statement which says two expressions can be true for a set of values assigned to the variable.

If a variable obeys two separate inequalities simultaneously, the set of inequalities is called a compound inequality. The simultaneous action may be a conjunction or may be a disjunction. If the set of inequalities are connected by the word ‘and’ in a compound inequality, then the set of inequalities mean a conjunction and if they are separated by the word ‘or’, then the set of inequalities mean a disjunction. When the set of inequalities have a conjunction then the solution must satisfy both the inequalities. If there is a disjunction, then the solution must satisfy one or both the inequalities.

The method of solving an inequality is exactly the same as solving an equation. All necessary algebraic operations can be performed as you do in case of equations for isolating the variable. There is only one very important difference to be noted. When you reverse the signs on both sides of an inequality, the inequality symbol also MUST be reversed.

A compound inequality is solved as per the following steps.

1) Solve for the variable in each inequality separately by necessary algebraic operations.

2) If the set of inequalities are in a conjunction, find the intersection of the set of the solutions from each case. In case they are in disjunction find the union of the intervals which can make both the inequality statements true.

3) Express the solution in interval notation or if specifically asked for, express in set builder notation.

4) Graph the solution on a number line in case of single variable or on a Cartesian grid in case of two variables.

Let us explain the concept and method to solve compound inequality with a few examples.

Example 1.

Solve the following compound inequality.

2x + 5 ≤ 11 and 1 – 3x < -2

First let us solve the inequality 2x + 5 ≤ 11

Subtracting 5 on both sides, 2x + 5 – 5 ≤ 11 – 5 or, 2x ≤ 6

now dividing both sides by 2, 2x ÷ 2 ≤ 6 ÷ 2 or, x ≤ 3

Now let us solve the inequality 1 – 3x < -2

Subtracting 1 on both sides, 1 – 3x – 1 < -2 – 1 or, -3x < -3

now dividing both sides by 3, -3x ÷ 3 < -3 ÷ 3 or, -x < -1

Changing the sign on both sides and reversing the inequality symbol, x > 1

The conclusion is ‘x’ can be of any value that is greater than 1 but less than or equal to 3.

Therefore, the solution is written in interval notation as, 1 < x ≤ 3

Now let us graph the solution of the compound inequality on a number line.

Since the solution lies between 1 and 3, shade the portion between 1 and 3. The value of ‘x’ does NOT include 1 whereas it includes 3. Hence put a open circle at 1 and a closed circle at x on the number line to highlight the inclusion and omission.

Example 2.

Solve the following compound inequality.

3x – 5 ≤ -14 or 2x + 1 > 9

First let us solve the inequality 3x – 5 ≤ -14

Adding 5 on both sides, 3x – 5 + 5 ≤ -14 + 5 or, 3x ≤ -9

now dividing both sides by 3, 3x ÷ 3 ≤ -9 ÷ 3 or, x ≤ -3

Now let us solve the inequality, 2x + 1 > 9

Subtracting 1 on both sides, 2x + 1 – 1 > 9 – 1 or, 2x > 8

now dividing both sides by 2, 2x ÷ 2 > 8 ÷ 2 or, x > 4

The conclusion is ‘x’ can be of any value that is less than or equal -3 OR greater than 4.

Therefore, the solution is written as a union in interval notation as, -∞ < x ≤ -3 U 4 < x < ∞

Now let us graph the solution of the compound inequality on a number line.

Since the value of ‘x’ can be anything less than or equal to -3, on a number line shade all the portion to the left of -3, with a closed circle at -3 meaning that point is included. Similarly since the value of ‘x’ can be anything greater than 4, on the same number line shade all the portion to the right of 4, with an open circle at 4 meaning that point is not included.

The two examples we illustrated may clearly explain the difference between the conjunction and disjunction in a compound inequality. In the first example, the conjunction stipulates that the value of the variable must be between 1 and 3, 3 included to make the statement of the compound inequality true. However, in the second example, because of the disjunction the compound inequality is true with an option of being ‘either’ x ≤ -3 ‘or’ x > 4.