Absolute value equations and inequalities

2.8  Absolute value equations and inequalities

  In this section we describe methods of solving equations and inequalities involving absolute value. Recall from Chapter 1 that the absolute value of a number a, written |a|, gives the distance from a to 0 on a number line. By this definition, the absolute value equation |x| = 3 can be solved by finding all real numbers at a distance of 3 units from 0. As shown in Figure 2.15, there are two numbers satisfying this condition, 3 and -3, so that the solution set of the equation |x| = 3 is the set {3,-3}.

  FIGURE 2.15

  ABSOLUTE VALUE EQUATIONS If a and b represent two real numbers, then the absolute value of their difference, |a-b| or |b-a|, represents the distance between the points on the number line whose coordinates are a and b. (Verify this for 3 and -3 in Figure 2.15.) This concept is used in simple equations involving absolute value.

Example 1.

USING THE DISTANCE DEFINITION TO SOLVE AN ABSOLUTE VALUE EQUATION

  Solve |p-4|=5.

  The expression |p-4| represents the distance between p and 4. The equation |p-4|=5 can be solved by finding all real numbers that are 5 units from 4. As shown in Figure 2.16, these numbers are -1 and 9. The solution set is {-1,9}.

  FIGURE 2.16

  The definition of absolute value leads to the following properties of absolute value that can be used to solve absolute value equations algebraically.

SOLVING ABSOLUTE VALUE EQUATIONS

  Solve each equation

  (a) |5-3m|=12

  Use property (1) above, with a=5-3m, to write

  5-3m=12  or  5-3m=-12.

  Solve each equation

  5-3m=12

  -3m=7

  m=-7/3

  or

  5-3m=-12

  -3m=-17

  m=17/3

  The solution set is {-7/3,17/3}.

  (b) |4m-3|=|m+6|

  By property (2) above, this equation will be true if

  4m-3=m+6  or  4m-3=-(m+6).

  Solve each equation.

  4m-3=m+6

  3m=9

  m=3

  or

  4m-3=-(m+6)

  4m-3=-m-6

  5m=-3

  m=-3/5

  The solution set of |4m-3|=|m+6| is thus {3,-3/5}.

  ABSOLUTE VALUE INEQUALITIES The method used to solve absolute value equations can be extended to solve inequalities with absolute value.

Example 3.

USING THE DISTANCE DEFINITION FOR ABSOLUTE VALUE INEQUALITIES

  (a) Solve |x|<5.

  Since absolute value gives the distance between a number and 0, the inequality |x|<5 is satisfied by all real numbers whose distance from 0 is less than 5. As shown in Figure 2.17, the solution includes all numbers from -5 to 5, or   . In interval notation, the solution is written as the open interval (-5,5). A graph of the solution set is shown in Figure 2.17.

  FIGURE 2.17

  (b) Solve |x|>5.

  In a manner similar to pan (a), we see that the solution of |x|>5 consists of all real numbers whose distance from 0 is greater than 5. This includes those numbers greater than 5 or those less than -5:x<-5 or x>5.

  FIGURE 2.18

  In interval notation, the solution is written (-∞,-5) {union} (5,∞). The solution set is shown in Figure 2.18. 

  The following properties of absolute value, which can be obtained from the definition of absolute value, are used to solve absolute value inequalities.

SOLVING AN ABSOLUTE VALUE INEQUALITIES

  For any positive number b

  1. |a|<b if and only if -b<a<b;

  2. |a|>b if and only if a<-b or a>b.

Example 4

SOLVING AN ABSOLUTE VALUE INEQUALITY

  Solve |x-2|<5.

  This inequality is satisfied by all real numbers whose distance from 2 is less than 5. As shown in Figure 2.19, the solution set is the interval (-3, 7). Property (1) above can be used to solve the inequality as follows. Let a=x-2 and b=5, so that |x-2|<5 if and only if

  -5<x-2<5.

  Adding 2 to each part of this three-pan inequality produces

  -3<x<7,

  giving the interval solution (-3,7).

  FIGURE 2.19

Example 5

SOLVING AN ABSOLUTE VALUE INEQUALITY

  Solve |x-8|>=1.

  All numbers whose distance from 8 is greater than or equal to 1 are solutions.
To find the solution using property (2) above, let a = x - 8 and b = 1 so that |x-8|>=1 if and only if

  x-8<=-1  or  x-8>=1

  x<=7  or  x>=9.

  The solution set, (-∞,7) {union} (9,∞), is shown in Figure 2.20

  FIGURE 2.20

  The properties given above for solving absolute value inequalities require that the absolute value expression be alone on one side of the inequality. Example 6 shows how to meet this requirement when this is not the case at first.

Example 6

SOLVING AN ABSOLUTE VALUE INEQUALITY REQUIRING A TRANSFORMATION

  Solve |2-7m|-1>4.

  In order to use the properties of absolute value given above, first add l to both sides; this gives 

  |2-7m|>5.

  Now use property (2) above. By this property, |2-7m|>5 if and only if

  2-7m<-5  or  2-7m>5.

  Solve each of these inequalities separately to get the solution set (-∞,-3/7) {union} (1,∞).

  If an absolute value equation or inequality is written with 0 or a negative number on one side. such as |2-5x|>=-4, we do not solve by applying the methods of the earlier examples. Use the fact that the absolute value of any expression must be a nonnegative number to solve the equation or inequality.

Example 7

SOLVING SPECIAL CASES OF ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

  Use the fact that absolute value is always inequality. ways nonnegative to solve each equation or inequality.

  (a) |2-5x|>=-4

  Since the absolute value of a number is always nonnegalive, |2-5x|>=-4 is always true. The solution set includes all real numbers, written (-∞,∞).

  (b) |4x-7|<-3

  The absolute value of any number will never be less than -3 (or less than any negative number). For this reason, the solution set of this inequality is {empty}.

  (c) |5x+15|=0

  The absolute value of a number will be zero only if that number is 0. There fore, this equation is equivalent to 5x+15=0, which has solution set {-3}.

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   We end this section with an example showing how certain statements involving distance can be described using absolute value inequalities.

Example 8

USING ABSOLUTE VALUE INEQUALITIES TO DESCRIBE DISTANCES

  Write each statement using an absolute value inequality.

  (a) k is not less than 5 units from 8.

  Since the distance from k to 8, written |k-8| or |8-k|. is not less than 5 the distance is greater than or equal to 5. Write this as

  |k-8|>=5.

  (b) n is within 0.001 of 6.

  This statement indicates that n may be 0.001 more than 6 or 0.001 less than 6. That is the distance of n from 6 is no more than 0.001, Written

  |n-6|<=0.001.