Basic Concepts Recall that the absolute value of a number a, written ∣ a ∣, gives the undirected distance from a to 0 on a number line. By this definition, the equation x = 3 can be solved by finding all real numbers at a distance of 3 units from 0. As shown in Figure 19, two numbers satisfy this equation, -3 and 3, so the solution set is 5-3, 36.
■ Basic Concepts
■ Absolute Value Equations
■ Absolute Value Inequalities
■ Special Cases
■ Absolute Value Models for Distance and Tolerance
0
– 3 3
Distance is
less than 3. Distance is greater than 3.
Distance is
greater than 3. Distance is less than 3.
Distance
is 3. Distance
is 3.
Figure 19
Similarly, x 63 is satisfied by all real numbers whose undirected dis- tances from 0 are less than 3. As shown in Figure 19, this is the interval
-36x63, or 1-3, 32.
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1.8 Absolute Value Equations and Inequalities
Finally, x 7 3 is satisfied by all real numbers whose undirected distances from 0 are greater than 3. These numbers are less than -3 or greater than 3, so the solution set is
1-∞, -32´13, ∞2.
Notice in Figure 19 that the union of the solution sets of x = 3, x 6 3, and x 7 3 is the set of real numbers.
These observations support the cases for solving absolute value equations and inequalities summarized in the table that follows. If the equation or inequal- ity fits the form of Case 1, 2, or 3, change it to its equivalent form and solve. The solution set and its graph will look similar to those shown.
Solving Absolute Value Equations and Inequalities Absolute Value
Equation or
Inequality* Equivalent Form
Graph of the
Solution Set Solution Set
Case 1: x =k x= k or x= -k –k k 5-k, k6
Case 2: x 6k -k6x6k
–k k 1-k, k2
Case 3: x 7k x6 -k or x7k
–k k 1-∞, -k2´ 1k, ∞2
*For each equation or inequality in Cases 1–3, assume that k70.
In Cases 2 and 3, the strict inequality may be replaced by its nonstrict form. Addi- tionally, if an absolute value equation takes the form a = b, then a and b must be equal in value or opposite in value.
Thus, the equivalent form of ∣a∣ = ∣b∣ is a =b or a = −b.
Absolute Value Equations Because absolute value represents undi- rected distance from 0 on a number line, solving an absolute value equation requires solving two possibilities, as shown in the examples that follow.
EXAMPLE 1 Solving Absolute Value Equations (Case 1 and the Special Case ∣ a ∣ = ∣ b ∣)
Solve each equation.
(a) 5- 3x = 12 (b) 4x- 3 = x+ 6 SOLUTION
(a) For the given expression 5- 3x to have absolute value 12, it must represent either 12 or -12. This equation fits the form of Case 1.
5- 3x = 12
5- 3x= 12 or 5 -3x = -12 Case 1 -3x= 7 or -3x = -17 Subtract 5.
x= -7
3 or x = 17
3 Divide by -3.
Check the solutions -73 and 173 by substituting them in the original absolute value equation. The solution set is E-73 , 173 F.
Don’t forget this second possibility.
(b) If the absolute values of two expressions are equal, then those expressions are either equal in value or opposite in value.
4x- 3 = x +6
4x- 3 =x + 6 or 4x -3 = -1x +62 Consider both possibilities.
3x =9 or 4x -3 = -x -6 Solve each linear equation.
x =3 or 5x= -3 x = -3 5
CHECK 4x - 3 = x+ 6 Original equation
P4A-35B - 3P≟P -35+ 6P Let x= - 3 5. 4132- 3 ≟ 3 +6 Let x=3.
P -125 -3P≟P -35 + 6P 12- 3 ≟ 3+ 6 P -275 P = P275 P ✓ True 9 = 9 ✓ True Both solutions check. The solution set is E-35 , 3F.
■✔ Now Try Exercises 9 and 19.
EXAMPLE 2 Solving Absolute Value Inequalities (Cases 2 and 3) Solve each inequality.
(a) 2x+ 1 67 (b) 2x+ 1 77 SOLUTION
(a) This inequality fits Case 2. If the absolute value of an expression is less than 7, then the value of the expression is between -7 and 7.
2x+ 1 67
-762x +1 67 Case 2
-86 2x 66 Subtract 1 from each part.
-46 x 63 Divide each part by 2.
The final inequality gives the solution set 1-4, 32 in interval notation.
(b) This inequality fits Case 3. If the absolute value of an expression is greater than 7, then the value of the expression is either less than -7 or greater than 7.
2x+ 1 77
2x+ 16 -7 or 2x + 177 Case 3
2x6 -8 or 2x76 Subtract 1 from each side.
x6 -4 or x73 Divide each side by 2.
The solution set written in interval notation is 1-∞, -42´ 13, ∞2.
■✔ Now Try Exercises 27 and 29.
Absolute Value Inequalities
LOOKING AHEAD TO CALCULUS The precise definition of a limit in calculus requires writing absolute value inequalities.
A standard problem in calculus is to find the “interval of convergence”
of a power series by solving the following inequality.
x-a6r
This inequality says that x can be any number within r units of a on the number line, so its solution set is indeed an interval—namely the interval
1a-r, a+r2.
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1.8 Absolute Value Equations and Inequalities
Cases 1, 2, and 3 require that the absolute value expression be isolated on one side of the equation or inequality.
EXAMPLE 3 Solving an Absolute Value Inequality (Case 3) Solve 2- 7x - 174.
SOLUTION 2- 7x - 17 4
2 -7x 7 5 Add 1 to each side.
2 -7x6 -5 or 2- 7x7 5 Case 3
-7x6 -7 or -7x7 3 Subtract 2 from each side.
x71 or x6 -3
7 Divide by -7. Reverse the direction of each inequality.
The solution set written in interval notation is A-∞, -37B ´ 11, ∞2.
■✔ Now Try Exercise 51.
Special Cases The three cases given in this section require the constant k to be positive. When k" 0, use the fact that the absolute value of any ex- pression must be nonnegative, and consider the conditions necessary for the statement to be true.
EXAMPLE 4 Solving Special Cases Solve each equation or inequality.
(a) 2- 5x Ú -4 (b) 4x -7 6 -3 (c) 5x +15 = 0 SOLUTION
(a) Since the absolute value of a number is always nonnegative, the inequality 2 -5x Ú -4 is always true.
The solution set includes all real numbers, written 1-∞, ∞2.
(b) There is no number whose absolute value is less than -3 (or less than any negative number).
The solution set of 4x- 7 6 -3 is ∅.
(c) The absolute value of a number will be 0 only if that number is 0. Therefore, 5x+ 15 = 0 is equivalent to
5x+ 15= 0, which has solution set 5-36. CHECK Substitute -3 into the original equation.
5x + 15 = 0 Original equation 51-32 + 15 ≟0 Let x= -3.
0= 0 ✓ True
■✔ Now Try Exercises 55, 57, and 59.
Absolute Value Models for Distance and Tolerance If a and b repre- sent two real numbers, then the absolute value of their difference,
either a- b or b- a, represents the undirected distance between them.
EXAMPLE 5 Using Absolute Value Inequalities with Distances Write each statement using an absolute value inequality.
(a) k is no less than 5 units from 8. (b) n is within 0.001 unit of 6.
SOLUTION
(a) Since the distance from k to 8, written k- 8 or 8- k, is no less than 5, the distance is greater than or equal to 5. This can be written as
k - 8 Ú5, or, equivalently, 8- k Ú 5. Either form is acceptable.
(b) This statement indicates that the distance between n and 6 is less than 0.001.
n- 6 6 0.001, or, equivalently, 6- n 6 0.001
■✔ Now Try Exercises 69 and 71.
EXAMPLE 6 Using Absolute Value to Model Tolerance
In quality control situations, such as filling bottles on an assembly line, we often wish to keep the difference between two quantities within some predetermined amount, called the tolerance.
Suppose y =2x +1 and we want y to be within 0.01 unit of 4. For what values of x will this be true?
SOLUTION y - 4 60.01 Write an absolute value inequality.
2x+ 1- 4 60.01 Substitute 2x+1 for y.
2x- 3 60.01 Combine like terms.
-0.0162x- 3 6 0.01 Case 2
2.996 2x 6 3.01 Add 3 to each part.
1.4956 x 6 1.505 Divide each part by 2.
Reversing these steps shows that keeping x in the interval 11.495, 1.5052 ensures that the difference between y and 4 is within 0.01 unit.
■✔ Now Try Exercise 75.
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1.8 Absolute Value Equations and Inequalities
CONCEPT PREVIEW Match each equation or inequality in Column I with the graph of its solution set in Column II.