§ 4.3 equations and inequalities involving absolute value

§ 4.3

Equations and Inequalities Involving Absolute Value

Blitzer, Intermediate Algebra, 5e – Slide #2 Section 4.3

Absolute Value Inequalities

We have taken the absolute value of numbers. We know that the absolute value of a number is its distance from 0 on the number line. We have not worked with equations or inequalities that contain absolute value.

By the end of this section, we will know how to solve equationsand inequalities containing absolute value and should also see the usefulness of this new tool.

To begin, consider x = 3

How many solutions does this equation have? Hint: How many numbers are exactly3 units away from 0 on the number line? What are the solutions?


Absolute Value Equations

Rewriting an Absolute Value Equation Without Absolute Value Bars

If c is a positive real number and X represents any algebraic expression, then |X| = c is equivalent to X = c or X = -c.

Rewriting an Absolute Value Equation with Two Absolute Values, Without Absolute Value Bars

If |X| = |Y|, then X = Y or X = -Y.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve: 3|y + 5| = 12.

We must first isolate the absolute value expression, |y + 5|.

Divide both sides by 3

Rewrite equation without absolute value bars

3|y + 5| = 12

|y + 5| = 4

y + 5 = 4 y + 5 = -4or

y = -1 y = -9 Subtract 5 from both sides

Now we will check the two solutions using the original equations.



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve: |3x - 5| = |3x + 5|.

We rewrite the equation without absolute value bars.

or3x - 5 = 3x + 5 3x - 5 = -(3x + 5)

We now solve the two equations that do not contain absolute value bars.

or3x - 5 = 3x + 5 3x - 5 = -(3x + 5)-5 = 5 3x - 5 = -3x - 5

6x - 5 = - 56x = 0

x = 0



Original equation

Substitute the proposed

3|y + 5| = 12

Simplify

The solutions are -1 and -9. We can also say that the solution set is {-1,-9}.

CONTINUECONTINUEDDCheck -1: Check -9:

3|y + 5| = 12

3|(-1) + 5| = 12 3|(-9) + 5| = 12

3|-1 + 5| = 12 3|-9 + 5| = 12

3|4| = 12 3|-4| = 12

3(4) = 12 3(4) = 12

12 = 12 12 = 12

Add

Simplify

Multiply

solutions

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truetrue



Since the first equation yielded -5 = 5, which is clearly a false statement, there is no solution for the first equation. However, the second equation yielded a potentially legitimate solution, x = 0. We now check it.

Check 0:

Original equation

CONTINUECONTINUEDD

|3(0) – 5| = |3(0) + 5| Replace x with 0|0 – 5| = |0 + 5| Multiply

|-5| = |5| Simplify5 = 5 Simplify

The solution is 0. We can also say that the solution set is {0}.

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|3x - 5| = |3x + 5|

true



We have now looked at solving absolute value equations. We have seen that “taking the bars off” in an absolute value equation usually leads us to write two separate equations to solve.

We have seen that we often get two distinct number solutions for these absolute value equations.

Now, we go forward to consider absolute value inequalities. As with the other inequalities we have solved, we now expect that our solutions will more often be intervals than points.



Solving an Absolute Value InequalityIf X is an algebraic expression and c is a positive number,

1) The solutions of |X| < c are the numbers that satisfy –c < X < c.

2) The solutions of |X| > c are the numbers that satisfy X < -c or X > c.

These rules are valid if < is replaced by and > is replaced by .

Absolute Value Inequalities with Unusual Solution Sets

If X is an algebraic expression and c is a negative number,

1) The inequality |X| < c has no solution.

2) The inequality |X| > c is true for all real numbers for which X is defined.

Blitzer, Intermediate Algebra, 4e – Slide #48


EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and graph the solution set on a number line: |x - 2| > 5.

We rewrite the inequality without absolute value bars. or x - 2 > 5

We now solve the compound inequality.

x - 2 < -5

or x - 2 > 5x - 2 < -5

x > 7x < -3 Add 2 to both sides



The solution set consists of all numbers that are less than -3 or greater than 7. The solution set is {x|x < -3 or x > 7}, or, in interval notation, . The graph of the solution set is shown as follows:

CONTINUECONTINUEDD

,73,

()

or x > 7x < -3

-3 7



An alternate method for solving absolute value inequalities involvesthe use of boundary points. See page 265 of your text for a full description of this method. A brief description of the process is:

First you replace the inequality symbol in your problem with an = and solve the equation involving the absolute value rather than the originalinequality. This gives you solutions which we call boundary points. These boundary points which work in the equation will separate what works from what fails to work in the inequality.

Choose any number that falls between the boundary points (called a test value) and see if that number works in your inequality. If so, the whole interval between the boundary points works. If not, no point in that interval works. Check each interval in this way. Finally, test your boundary points to see if they work in the inequality so that you will know whether to use open or closed intervals in your solution.

Blitzer, Intermediate Algebra, 4e – Slide #48


EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use the boundary point method to solve and graph the problem we previously worked: |x - 2| > 5.

To find the boundary points, we write the equation

We now solve this equation. The two equations that are generated are:

or x - 2 = 5x - 2 = -5

x = 7x = -3 The two boundary points are -3 and 7.

|x - 2| = 5



These two points divide the line into three intervals and we should check each one. We can check any number less than -3. Let’s choose -5 for a test point. We put -5 into the original inequality to see if it works. It makes the inequality say “7 > 5” which is true. That interval works. Check 0 in the next interval. Put it into the original inequality. It makes the statement “2 > 5” which is false. That interval doesn’t work. Check any number in the last interval. Let’s check 10. When we plug in 10, we get the sentence “8 > 5” which is true. Then the last interval of numbers greater than 7 works. See the solution below.

CONTINUECONTINUEDD

,73,

()

x < -3

-3 7

x > 7or

Note that the boundary points of -3 and 7 don’t work here since the original inequalityrequired an absolute value strictly greater than 5. The boundary points made the absolute value exactly equal to five.

Solution: Interval notation:



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Solve and graph the solution set on a number line: .

We rewrite the inequality without absolute value bars.

We now solve the compound inequality.

Multiply

20213 x

2021320 x

201320 x

2023320 x

21319 x

73

19

x

Simplify

Add 1 to all three sides

Divide all three sides by 3



The solution set is all real numbers greater than or equal to and less than or equal to 7, denoted by or [ ,7]. The graph of the solution set is as follows.

7

3

19| xx

3

19

CONTINUECONTINUEDD

3

193

19

- ]

[

73

19

x

7



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

The specifications for machine parts are given with tolerance limits that describe a range of measurements for which the part is acceptable. In this example, x represents the length of a machine part, in centimeters. The tolerance limit is 0.01 centimeters. Solve: . If the length of the machine part is supposed to be 9.4 centimeters, interpret the solution.

01.04.9 x

01.04.9 x

01.04.901.0 x

Add 9.4 to all three sides

Original inequality

Rewrite inequality

41.939.9 x

First we solve the given inequality.



. 419399or 41.939.9| .,.xx Therefore, the solution set is So, a machine part that is supposed to be 9.4 centimeters is acceptable between a low of 9.39 centimeters and a high of 9.41 centimeters.

CONTINUECONTINUEDD


Important to remember…

When you solve an absolute value inequality, your answer will probably be an interval of values rather than a single number. Using boundary points to find the solution of an absolute value inequality is usually a quick and easy method to use. Find points on the “edge” of the solution, that is those points that work when you substitute an = for the inequality symbol. Consider the intervals that these boundary points divide the line into. Check a number within each interval. If that number works in the original inequality, the whole interval does. After you find which intervals work, you must then see which boundary points work. This will determine whether your interval solutions are closed (containing both endpoints), open (containing neither endpoint), or half open or closed.

§ 4.3 equations and inequalities involving absolute value

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