Intermediate Algebra/Solving Inequalities

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Inequalities[edit]

As opposed to an equation, an inequality is an expression that states that two quantities are unequal or not equivalent to one another. In most cases we use inequalities in real life more than equations (i.e. this shirt costs $2 more than that one).

Given that a and b are real numbers, there are four basic inequalities:

a < b
a is "less than" b:
Example: 2 < 4 ; -3 < 0; etc.
a > b
a is "greater than" b:
Example: -2 > -4 ; 3 > 0 ; etc.
a \le b
a is "less than or equal to" b:
Example: If we know that x \le 7, then we can conclude that x is equal to any value less than 7, including 7 itself.
a \ge b
a is "greater than or equal to" b:
Example: Conversely, if x \ge 7, then x is equal to any value greater than 7, including 7 itself.

Properties of Inequalities[edit]

Just as there are four properties of equality, there are also four properties of inequality:

Addition Property of Inequality

If a, b, and c are real numbers such that a > b, then a + c > b + c. Conversely, if a < b, then a + c < b + c.

Subtraction Property of Inequality

If a, b, and c are real numbers such that a > b, then a - c > b - c. Conversely, if a < b, then a - c < b - c.

Multiplication Property of Inequality

If a, b, and c are real numbers such that a > b and c > 0, then ac (or a * c) > bc (or b * c). Conversely, if a < b and c > 0, then ac < bc. (Note that if c = 0, then both sides of the inequality are in fact equal.) We will also review cases where c is less than zero later in the lesson.

Division Property of Inequality

If a, b, and c are real numbers such that a > b, and c > 0, then \frac {a}{c} > \frac {b}{c}. Conversely under the same conditions, if a < b, then \frac {a} {c} < \frac {b}{c}. As with the Multiplication Property, there are special cases that will be discussed later when c < 0.


Note that all four properties also work with inequalities where a \le b or a \ge b.

Trichotomy Property[edit]

The statements above form the basis of Trichotomy Property:

Given any two real numbers a and b, then only one of the following statements must hold true:

  1. a < b
  2. a = b
  3. a > b

So, if we are given any two unknown real-number values, then any one of the three statements will hold true.

Solving Inequalities[edit]

Solving algebraic inequalities is more or less identical to solving algebraic equations. Consider the following example:

                                2x + 4 \le 3x - 7

Although it may be an inequality, we can use the Properties of Inequality stated above to solve. Start by subtracting 2x from both sides:

                                2x - 2x + 4 \le 3x - 2x - 7
                                4 \le x - 7

Finish by adding 7 to both sides.

                                4 + 7 \le x - 7 + 7 
                                11 \le x

This can be rewritten as x \ge 11. To check, substitute any value greater than or equal to 11. However, in order to satisfy the Trichotomy Property, we'll substitute three different values: 10, 11, and 12.

2x + 4 \le 3x - 7 2x + 4 \le 3x - 7 2x + 4 \le 3x - 7
2(10) + 4 \le 3(10) - 7 2(11) + 4 \le 3(11) - 7 2(12) + 4 \le 3(12) - 7
24 \le 23 26 \le 26 28 \le 29

Ten is incorrect, whereas eleven and twelve satisfy the solution. Therefore, the solution set - the set of all answers which satisfy the original inequality - is x \ge 11. Written in set notation, the answer is {x | x \ge 11}. This is read as "the set of all x such that x is greater than or equal to 11".

Special Cases - A variable in the denominator[edit]

For example consider the inequality

\frac{2}{x-1}<2\,

In this case one cannot multiply the right hand side by (x-1) because the value of x is unknown. Since x may be either positive or negative, you can't know whether to leave the inequality sign as <, or reverse it to >. The method for solving this kind of inequality involves four steps:

  1. Find out when the denominator is equal to 0. In this case it's when x=1.
  2. Pretend the inequality sign is an = sign and solve it as such: \frac{2}{x-1}=2\,, so x=2.
  3. Plot the points x=1 and x=2 on a number line with an unfilled circle because the original equation included < (it would have been a filled circle if the original equation included <= or >=). You now have three regions: x<1, 1<x<2, and x>2.
  4. Test each region independently. in this case test if the inequality is true for 1<x<2 by picking a point in this region (e.g. x=1.5) and trying it in the original inequation. For x=1.5 the original inequation doesn't hold. So then try for 1>x>2 (e.g. x=3). In this case the original inequation holds, and so the solution for the original inequation is 1>x>2.



Practice Problems

Determine if the number in parentheses satisfies the given inequality:

1. x > 4 (5)

2. 2x \le 9 (6.5)

3. 4x - 2 > 8 (2.5)


Answers

1. Yes; 5 is greater than 4.

2. No; 2(6.5) = 13, and 13 is neither less than nor equal to 9.

3. No; although substituting 2.5 for x makes both sides equal, the inequality states that 4x - 2 must be greater than 8.

Special Cases[edit]

Let us suppose that we were given this inequality to solve:

                                -3x + 8 \ge 26

Using the same steps as above, start by subtracting 8 from both sides.

                                -3x + 8 - 8 \ge 26 - 8
                                -3x \ge 18

Now divide both sides by -3.

                                -3x/-3 \ge 18/-3
                                x \ge -6

Check this solution by substituting three numbers. We'll use -7, -6, and -5.

-3x + 8 \ge 26 -3x + 8 \ge 26 -3x + 8 \ge 26
-3(-7) + 8 \ge 26 -3(-6) + 8 \ge 26 -3(-5) + 8 \ge 26
29 \ge 26 26 \ge 26 23 \ge 26

Wait, what happened? -7 and -6 satisfy the inequality, yet -7 is non-inclusive in the solution set! And -5, which is greater than -6, does not satisfy the inequality!

This is a special case involved in solving inequalities. Because the coefficient of the x-term was negative, the constant on the other side (26, which became 18 and then -6) switched signs. In order to attain a valid solution if a negative number is divided, we need to switch the sign in order to make sure that the solution set is correct. Thus, x \ge -6 becomes x \le -6, or more specifically, {x | x \le -6}.


Practice Problems

Solve the following inequalities and check each solution set:


1. -3x - 5 > 22

2. 4x - 2 \le -6x - 7

3. \frac {-2x}{5} \ge 12 - 4x

4. 11 - 7x < 39


Answers

1. x < -9

2. x \le \frac {-1}{2}

3. x \ge \frac {10}{3}

4. x > -4

Graphing Solutions[edit]

Because inequalities have multiple solutions, we need to be able to represent them graphically. In order to do so, we use the number line, depicted below:


                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

There are two ways of graphing solutions; however, each one is unique depending on the nature of the inequality. If, for example, the inequality contains the < or > sign, we use an open circle ("O") and place it on (or above) the corresponding position on the number line, then draw another line either left or right of the solution (based on the sign) to indicate the infinite number of solutions in the set. For example, if we were to graph x < 4:


                        <-----------------------------O
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

The open circle indicates that 4 is not included in the solution set; however, all values less than 4 satisfy the solution.

If an inequality contains a \le or \ge, then a closed circle (it will be depicted here with a *) is placed on or above the corresponding position on the number line. Hence, the solution x \ge -2 is graphed as follows.

                                    *----------------------->
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

The asterisk (*) indicates that -2 is inclusive in the solution set, as are all values greater than -2.

Practice Problems

Graph the following inequalities on a number line:

1. x > 4

2. x \le 1

3. -1 > x

Solutions

1.

                                                      O----->
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

2.

                        <--------------------*
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

(Note that your actual graph should have a closed circle in place of the dot. If in doubt, simply draw a circle and color it black.)

3. Be careful here; you need to rearrange the inequality first before graphing. When rewritten, the inequality becomes x < -1:

                        <--------------O
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5


Lesson Review[edit]

An inequality is a statement justifying that two quantities are not equal to each other. There are four cases of inequalities, two of which allow for equality (a \le b and a \ge b). The four properties of inequality, which are more or less parallel to the properties of equality, can be used to solve simple inequalities. The only exceptions are in multiplication and division, where the signs must be reversed if both sides are multiplied or divided by a negative number. Finally, the solution set of an inequality can be graphed on the number line with either an open circle ("O") or a closed circle ("*"), depending on the original sign used.

Lesson Quiz[edit]

1. What is an inequality? List the four possible cases of inequalities.

2. State the Trichotomy Property in your own words.

3. Solve the following and graph all solutions:

  a. 4x + 3 > 7                                b. 2x - 4 \le -x + 1
  c. \frac {-12c}{3} \ge -24      
  d. -x + 4 > 3x

Quiz Answers[edit]

1. An inequality states that two quantities are not equal. The four cases of inequalities: a < b, a > b, a \le b, a \ge b.

2. Answers will vary. Just make sure you didn't copy and paste the actual definition.

3a. x > 1

                                             O-------------->
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

3b. x \le \frac {5}{3}

                        <----------------------*
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5

3c. x \le 6

                     <--------------------------------------*
                     <--|--|--|--|--|--|--|--|--|--|--|--|--|--> 
                       -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6

3d. x < 1

                        <--------------------O
                        <--|--|--|--|--|--|--|--|--|--|--|-->
                          -5 -4 -3 -2 -1  0  1  2  3  4  5