Properties of Inequalities

1. Addition and Subtraction Properties

If a < b, then a + c < b + c

If a < b, then a – c < b – c

The same quantity may be added to or subtracted from both sides of an inequality without changing the solution set

2. Positive Multiplication and Division Properties

If a < b and c is positive, then a · c < b · c

If a < b and c is positive, then <

Both sides of an inequality may be multiplied or divided by the same positive number without changing the solution set.

3. Negative Multiplication and Division Properties

If a < b and c is negative, then a · c > b · c

If a < b and c is negative, then >

Both sides of an inequality may be multiplied or divided by the same negative number without changing the solution set as long as the direction of the inequality symbol is reversed.

REMEMBER: Always reverse the direction of the inequality symbol when multiplying or dividing by a negative number.

Example. Solve: 7(x + 4) – 13 > 12 + 13(3 + x)

Solution:

7(x + 4) – 13 > 12 + 13(3 + x)
7x + 28 – 13 > 12 + 39 + 13x	Distributive Property
7x + 15 > 13x + 51	Combine like terms
7x + 15 – 13x > 13x + 51 – 13x	Subtract 13x from both sides
-6x + 15 > 51	Combine like terms
-6x + 15 – 15 > 51 – 15	Subtract 15 from both sides
-6x > 36	Combine like terms
<	Divide both sides by –6 and reverse the direction of the inequality symbol
x < -6	Simplify

The solution set consists of all real numbers that are greater than –6. Interval notation for the solution set is (-¥ ,-6].

Example. Solve: 1 < < 9

Solution.

To solve this three-part inequality, apply the properties of inequalities, working with three parts of the inequality at the same time. The goal is to isolate the variable in the middle part.

1 < < 9
-2 · 1 > -2 · > -2 · 9	Multiply all three parts of the inequality by –2; reverse the direction of both inequality symbols
-2 > 4m – 5 > -18	Simplify
-2 + 5 > 4m – 5 + 5 > -18 + 5	Add 5 to all three parts of the inequality
3 > 4m > -13	Simplify
> >	Divide all three parts by 4. Since we are dividing by a positive number, the inequality symbols are not reversed.
> m >	Simplify

The solution set consists of all real numbers between and . Interval notation for this solution set is .

Quadratic Inequalities

A quadratic inequality is an inequality that can be written in the form ax + bx + c < 0, where a, b, and c are real numbers, and c ¹ 0. (The symbol < can be replaced with >, <, or >.)

Procedure for Solving Quadratic Inequalities

1. Write the inequality in standard form: ax + bx + c < 0. ( The symbol might also be >, <, or >.)

2. Replace the inequality symbol with =, and solve the equation ax + bx + c = 0.

3. Locate the solutions obtained in step 2 on a number line; the solution(s) divide the number line into test intervals.

4. Choose one number from each test interval, and substitute it into the original inequality. If the substitution produces a true statement, then all real numbers in the test interval are in the solution set. If the substitution produces a false statement, then the real numbers in the test interval are not in the solution set.

Example. Solve: 2x + x > 15

Solution.

2x + x > 15
2x + x - 15 > 0	Subtract 15 from both sides to obtain standard form
2x + x - 15 = 0	Replace > with =
(2x – 5)(x + 3) = 0 2x – 5 = 0 or x + 3 = 0 2x = 5 or x = -3 x =	Solve by factoring

The solutions obtained above divide the number line into three test intervals:

(-¥ , -3) , (-3, ) , (, ¥ ).

We will take one number from each test interval and substitute that number into the original inequality:

(-¥ , -3) Test number: -4 (Any number less than –3 could be chosen)

2x + x > 15 Original inequality

2(-4)+(-4) > 15 Substitution of –4

2(16) – 4 > 15

32 – 4 > 15

28 > 15 True

Conclusion: (-¥ , -3) belongs to the solution set.

(-3, ) Test number: 0

2x + x > 15

2(0)+ (0) > 15

0 + 0 > 15

0 > 15 False

Conclusion: (-3, ) does not belong to the solution set.

(, ¥ ) Test number: 3

2x + x > 15

2(3) + 3 > 15

2(9) + 3 > 15

18 + 3 > 15

21 > 15 True

Conclusion: (, ¥ ) belongs to the solution set.

Thus, the solution set is (-¥ , -3) È (, ¥ )

Rational Inequalities

A rational inequality involves quotients of algebraic expressions.

Procedure for Solving Rational Inequalities

1. Write the inequality so that one side is 0.

2. If necessary, combine terms on the other side into a single quotient.

3. Find the number(s) which make the numerator and denominator equal to zero.

4. Locate the numbers obtained in step 3 on the number line; these numbers divide the number line into test intervals.

5. Choose one number from each test interval and substitute it into the original inequality. If the substitution produces a true statement, then all real numbers in the test interval are in the solution set. If the substitution produces a false statement, then the real numbers in the test interval are not in the solution set.

Example. Solve:

Solution.

	Subtract 3 from both sides to get 0 on one side
	Get a common denominator; subtract fractions to obtain a single quotient
7 – 5x = 0 or 2x – 1 = 0 7 = 5x or 2x = 1 or x =	Find numbers which make the numerator and the denominator equal to 0; solve each equation

The solutions obtained above divide the number line into three test intervals:

(-¥ ,) , (, ) , (, ¥ )

We will take one number from each test interval and substitute it into the original inequality:

(-¥ ,) Test number: 0

Original inequality

-4 > 3 False

Conclusion: (-¥ ,) does not belong to the solution set.

(, ) Test number: 1

Original inequality

5 > 3 True

Conclusion: (, ) belongs to the solution set.

(, ¥ ) Test number: 2

Original inequality

2 > 3 False

Conclusion: (, ¥ ) does not belong to the solution set.

The interval : (, ) is in the solution set. Because the inequality symbol is >, we must include values which make the expression equal to 0. This would be the value which makes the numerator, 7 – 5x, equal to 0; that number is . Thus, the solution set is the interval (, ].