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Solving Absolute Value Inequalities
Solve the inequalities.
a. b.
Solution: a. Isolate the absolute value first. The inequality is in the form , where
Rewrite in the equivalent form Solve for w.
The solution is or, equivalently in interval notation, 1 4, 103 2.
5 w 0 4 6 w 6 103 6
4 6 w 6
12 6 3 w 6 10
11 6 3 w 1 6 11 a 6 x 6 a.
x 3 w 1.
0 x 0 6 a
03 w 1 0 6 11
03 w 1 0 4 6 7
3 1 `
03 w 1 0 4 6 7 t 5 `
Example 1
3. Solution:
3 6 x 6 3 The set of all points less than 3 units from zero
0 x 0 6 3
1 2 3 4 5 6
3 units
6 5 4 3 2 1 0
3 units
6 5 4 3 2 1 0 1 2 3 4 5 6 10 3
Absolute Value Equations and Inequalities
Let a be a real number such that Then Equation/ Solution Inequality (Equivalent Form) Graph
0 x 0 6 a a 6 x 6 a
0 x 0 7 a x 6 a or x 7 a
0 x 0 a x a or x a
a 7 0.
a a
a a
a a
To solve an absolute value inequality, first isolate the absolute value and then rewrite the absolute value inequality in its equivalent form.
650 Chapter 9 More Equations and Inequalities
1 2 3 4 5 6
3 units
6 5 4 3 2 1 0
3 units
Solution:
The set of all points more than 3 units from zero
x 6 3 or x 7 3
0 x 0 7 3
Section 9.4 Absolute Value Inequalities 651
Calculator Connections
Graph and On the given display window, ( Y 1 is below Y 2 ) for 4 6 x 6 103.
Y 1 6 Y 2
Y 1 abs 13 x 12 4 Y 2 7.
( ( 4
Y 1 0 3 x 1 0 4
(4, 7)
Y 2 7
(^103)
(^10 3 , 7 )
b.
Write the inequality with the absolute value on the left.
Isolate the absolute value. The inequality is in the form , where
Rewrite in the equivalent form
Solve the compound inequality.
Clear fractions.
The solution is or, equivalently in interval notation, 1 , 6 4 ´ 3 14, 2.
5 t 0 t 6 or t 146
t 6 or t 14
2 a
tb 2132 or 2 a
tb 2172
t 3 or
t 7
x a or x a.
t 5 2 or
t 5 2
0 x 0 a x 12 t 5.
`
t 5 ` 2
1 `
t 5 ` 3
3 1 `
t 5 `
6 14
Graph and On the given display window, for x 6 or x 14.
Y 1 Y 2
Y 1 abs 111 22 x 52 1 Y 2 3.
Calculator Connections
6 14
(6, 3) (14, 3) Y 2 ^3
Y 1 0 12 x 5 0 1
Solve the inequalities. Write the solutions in interval notation.
1. 2.
By definition, the absolute value of a real number will always be nonnega- tive. Therefore, the absolute value of any expression will always be greater than
5 6 1 `
ƒ 2 t 5 ƒ 2 11 c 1 `
Skill Practice
Skill Practice Answers 1.
2. 1 , 92 h 1 15, 2
3 7, 2 4
Section 9.4 Absolute Value Inequalities 653
1 0 1 2 3 4 5 6
6 5 4 3 2 1 2
2. Solving Absolute Value Inequalities by
the Test Point Method
For each problem in Example 1, the absolute value inequality was converted to an equivalent compound inequality. However, sometimes students have difficulty setting up the appropriate compound inequality. To avoid this problem, you may want to use the test point method to solve absolute value inequalities.
Calculator Connections
Graph From the graph, at (the x -intercept). On the given display window, for x 6 ^12 or x 7 ^12.
Y 1 7 0
Y 1 0 x ^12
Y 1 abs 14 x 22.
^12
Y 1 0 4 x 2 0
Solve the inequalities.
5. ƒ 3 x 1 ƒ 0 6. ƒ 3 x 1 ƒ 7 0
Skill Practice
Skill Practice Answers 5.
6. a,^1 3 b h a^1 3 e x | x , b 1 3 f ;
1 , 2
The second equation is the same as the first.
Therefore, exclude from the solution.
The solution is or equivalently in interval notation, 1 , ^12 2 ´ 1 ^12 , 2.
5 x 0 x ^12
x x ^12
4 x 2
4 x 2 0 or 4 x 2 0
04 x 2 0 0
Solving Inequalities by Using the Test Point Method
1. Find the boundary points of the inequality. (Boundary points are the real solutions to the related equation and points where the inequality is unde- fined.) 2. Plot the boundary points on the number line. This divides the number line into regions. 3. Select a test point from each region and substitute it into the original inequality. - If a test point makes the original inequality true, then that region is part of the solution set. 4. Test the boundary points in the original inequality. - If a boundary point makes the original inequality true, then that point is part of the solution set.
26 CHAPTER 1 Equations and Inequalities 1-
B. Solving “Less Than” Absolute Value Inequalities Absolute value inequalities can be solved using the basic concept underlying the property of absolute value equalities. Whereas the equation asks for all numbers x whose distance from zero is equal to 4, the inequality asks for all numbers x whose distance from zero is less than 4.
As Figure 1.6 illustrates, the solutions are and , which can be written as the joint inequality This idea can likewise be extended to include the absolute value of an algebraic expression X as follows.
Property I: Absolute Value Inequalities
If X represents an algebraic expression and k is a positive real number, then implies
EXAMPLE 4 ^ Solving “Less Than” Absolute Value Inequalities Solve the inequalities:
a. b.
Solution ^ a. original inequality
multiply by 4 apply Property I subtract 2 from all three parts
divide all three parts by 3
The solution interval is
b. original inequality Since the absolute value of any quantity is always positive or zero, the solution for this inequality is the empty set: { }.
Now try Exercises 25 through 38
C. Solving “Greater Than” Absolute Value Inequalities For “greater than” inequalities, consider Now we’re asked to find all numbers x whose distance from zero is greater than 4. As Figure 1.7 shows, solutions are found in the interval to the left of 4,or to the right of 4. The fact the intervals are disjoint
x (^7) 4.
2 x 7 6 5
2 x
6 3 x 2
4 3 x 2 4
3 x 2 (^) 4
3 x 2 4
2 x 7 (^6) 5
3 x 2 4
k 6 X 6 k
X 6 k
4 6 x 6 4.
x 7 4 x 6 4
0
Distance from zero is less than 4
) ) 5 4 3 2 1 1 2 3 4 5
x 6 4
x (^) 4
B. You’ve just learned how to solve less than absolute value inequalities
WORTHY OF NOTE
As with the inequalities from Section 1.2, solutions to absolute value inequalities can be checked using a test value. For Example 4(a), substituting from the solution interval yields:
✓ 1 2 1
x 0
Figure 1.
WORTHY OF NOTE
Property I can also be applied when the “ ” symbol is used. Also notice that if the solution is the empty set since the absolute value of any quantity is always positive or zero.
k 6 0,
402 Chapter 4 Absolute Value Functions
Version: Fall 2007
And we can describe the solution using interval and set-builder notation.
( −∞, 1 / 2) ∪ (1 , ∞ ) = {x : x < 1 / 2 or x > 1 }
Again, let a > 0. As we did with |x| ≤ a , we can take the union of the solutions of
|x| = a and |x| > a to find the solution of |x| ≥ a. This leads to the following property.
Property 19. If a > 0, then the inequality |x| ≥ a is equivalent to the inequality
x ≤ −a or x ≥ a.
I Example 20. Solve the inequality 3 | 1 − x| − 4 ≥ | 1 − x| for x.
Again, at first glance, the inequality
3 | 1 − x| − 4 ≥ | 1 − x|
looks unlike any inequality we’ve attempted to this point. However, if we subtract
| 1 − x| from both sides of the inequality, then add 4 to both sides of the inequality, we
get
3 | 1 − x| − | 1 − x| ≥ 4.
On the left, we have like terms. Note that 3 | 1 −x|−| 1 −x| = 3 | 1 −x|− 1 | 1 −x| = 2 | 1 −x|.
Thus,
2 | 1 − x| ≥ 4.
Divide both sides of the last inequality by 2.
| 1 − x| ≥ 2
We can now use Property 19 to write
1 − x ≤ − 2 or 1 − x ≥ 2.
We can solve each of these inequalities independently. First, subtract 1 from both sides
of each inequality, then multiply both sides of each resulting inequality by − 1, reversing
each inequality as you go.
1 − x ≤ − 2 or 1 − x ≥ 2
− x ≤ − 3 − x ≥ 1
x ≥ 3 x ≤ − 1
We prefer to write this in the order
x ≤ − 1 or x ≥ 3.
Section 4.4 Absolute Value Inequalities 403
Version: Fall 2007
We can sketch the solutions on a number line.
And we can describe the solutions using interval and set-builder notation.
( −∞, − 1] ∪ [3 , ∞ ) = {x : x ≤ − 1 or x ≥ 3 }
Revisiting Distance
If a and b are any numbers on the real line, then the distance between a and b is found
by taking the absolute value of their difference. That is, the distance d between a and
b is calculated with d = |a − b|. More importantly, we’ve learned to pronounce the
symbolism |a − b| as “the distance between a and b .” This pronunciation is far more
useful than saying “the absolute value of a minus b .”
I Example 21. Solve the inequality |x − 3 | < 8 for x.
This inequality is pronounced “the distance between x and 3 is less than 8.” Draw
a number line, locate 3 on the line, then note two points that are 8 units away from 3.
Now, we need to shade the points that are less than 8 units from 3.
Hence, the solution of the inequality |x − 3 | < 8 is
( − 5 , 11) = {x : − 5 < x < 11 }.
I Example 22. Solve the inequality |x + 5 | > 2 for x.
First, write the inequality as a difference.
|x − ( − 5) | > 2
This last inequality is pronounced “the distance between x and − 5 is greater than 2.”
Draw a number line, locate − 5 on the number line, then note two points that are 2
units from − 5.
Solution
Given: We are given an absolute value inequality, ||x+3| - 12| < 13 Note that we do not have any constraint on x. So, x can be an integer or a decimal
To find: We need to find the range of values of x, that satisfies the given inequality
Approach and Working: First, let us substitute the inner modulus function, |x + 3|, with t, which gives, |t - 12| < 13 We know that, the range of x, for |x| < a, is a < x < -a So, the range of x, for which |t - 12| < 13, is -13 < t – 12 < 13, Adding 12 on all the sides, we get, -1 < t < 25 Now, substituting, back the value of t as |x + 3|, we get, -1 < |x + 3| < 25 Since, the value of |x + 3| is always greater than or equal to 0 Hence, 0 ≤ |x + 3| < 25 Now, solving the inequality, |x + 3| < 25, we get the range of x as (-28, 22)
Therefore, the range of x, that satisfies the inequality, ||x+3| - 12| < 13, is (-28, 22)
Hence, the correct answer is option C.
https://gmatclub.com/forum/if-x-3-12-13-what-is-the-range-of-x-276038.html
- To solve our last inequality, , we re-write the absolute value using cases. For , , so we get , or. Finding the zeros of , we get and. However, we are only concerned with the portion of the number line where , so the only zero that we concern ourselves with is. This divides the interval into two intervals: and. We choose and as our test values. We find and. Hence, our solution to for is. Next, we turn our attention to the case. Here, , so our original inequality becomes , or. Setting , we find the zeros of to be and. Of these, only lies in the region , so we ignore. Our test intervals are now and. We choose and as our test values and find and. Hence, our solution to , in this region is.
Combining these into one sign diagram, we have that our solution is. Graphically, to check , we set and and look for the values where the graph of is above the the graph of (the solution of ) as well as the -coordinates of the intersection points of both graphs (where ). The combined sign chart is given on the left and the graphs are on the right.
2 x − x^2 ≥ | x − 1 | − 1
x < 1 | x − 1 | = −( x − 1 ) = 1 − x 2 x − x^2 ≥ 1 − x − 1 x^2 − 3 x ≤ 0
f ( x ) = x^2 − 3 x x = 0 x = 3
x < 1
x = 0 x < 1
(−∞, 0 ) ( 0 , 1 ) x = − 1 x = 12 f (− 1 ) = 4
f ( 12 ) = −^54 x^2 − 3 x ≤ 0 x < 1 [ 0 , 1 )
x ≥ 1 | x − 1 | = x − 1
2 x − x^2 ≥ x − 1 − 1 x^2 − x − 2 ≤ 0 g ( x ) = x^2 − x − 2 g
x = − 1 x = 2 x = 2 x ≥ 1 x = − 1
[ 1 , 2 ) ( 2 , ∞) x = 1 x = 3
g ( 1 ) = − 2 g ( 3 ) = 4 g ( x ) = x^2 − x − 2 ≤ 0
[ 1 , 2 )
[ 0 , 2 ]
2 x − x^2 ≥ | x − 1 | − 1 h ( x ) = 2 x − x^2 i ( x ) = | x − 1 | − 1 x
h i h ( x ) > i ( x )
x h ( x ) = i ( x )
https://math.libretexts.org/Bookshelves/Precalculus/Book:Precalculus(StitzZeager)/2:_Linear_and_Quadr...
| x^2 − 3 x + 1 | < 1 ⟺ − 1 < x^2 − 3 x + 1 < 1
We have, using the lower bound,
− 1 < x^2 − 3 x + 1 ⟺ x^2 − 3 x + 2 > 0 ⟺ ( x − 2 )( x − 1 ) > 0 ⟺ x ∈ (−∞, 1 ) ∪ ( 2 , ∞)
Also, we have, using the upper bound,
x^2 − 3 x + 1 < 1 ⟺ x ( x − 3 ) < 0 ⟺ x ∈ ( 0 , 3 )
For both upper and lower bounds to hold, we need the intersection of these solution sets which is
( 0 , 1 ) ∪ ( 2 , 3 ).
https://math.stackexchange.com/questions/1342995/quadratic-absolute-value-inequality
We need to solve the following system
and
≤ 1
x^2 − 5 x + 4 x^2 − 4
For the first we need to solve
or
which by the intervals method gives or.
≥ − 1
x^2 − 5 x + 4 x^2 − 4
− 1 ≤ 0
x^2 − 5 x + 4 x^2 − 4
≤ 0 ,
8 − 5 x ( x − 2 )( x + 2 )
x > 2 − 2 < x ≤ 1.
The second inequality gives
or
or
which by the intervals method again gives or or.
x^2 − 5 x + 4 x^2 − 4
≥ 0
2 x^2 − 5 x x^2 − 4
≥ 0 ,
x ( 2 x − 5 ) ( x − 2 )( x + 2 )
x ≥ 2.5 0 ≤ x < 2 x < − 2
Thus, after solving of this system we'll get the answer:
[ 0 , 1.6] ∪ [2.5, +∞)
https://math.stackexchange.com/questions/2339643/solve-absolute-value-inequation-with-fraction
Now I'll do the "plus" case:
x – 1 = +
x = 4
I've considered every case, and have arrived at four solutions. My complete answer is:
x = –2, 0, 2, 4
https://www.purplemath.com/modules/solveabs2.htm
We have
| x | + | x + 2 | =
4 − 3 p 1 − p
(1)
Case 1 : For ,
Here,.
x > 0
( 1 ) ⟺ x + ( x + 2 ) = ⟺ x =
4 − 3 p 1 − p
2 − p 2 ( 1 − p )
0 ⟺ p < 1 or p > 2
2 − p 2 ( 1 − p )
Case 2 : For − 2 < x ≤ 0 ,
( 1 ) ⟺ − x + ( x + 2 ) = ⟺ p = 2
4 − 3 p 1 − p
Case 3 : For ,
Here,.
x ≤ − 2
( 1 ) ⟺ − x − ( x + 2 ) = ⟺ x =
4 − 3 p 1 − p
6 − 5 p 2 ( p − 1 )
≤ − 2 ⟺ p < 1 or p ≥ 2
6 − 5 p 2 ( p − 1 )
So, the answer is the followings :
For p < 1 or p > 2 , x = ,.
2 − p 2 ( 1 − p )
6 − 5 p 2 ( p − 1 )
For p = 2 , − 2 ≤ x ≤ 0.
For 1 ≤ p < 2 , there is no such x.
https://math.stackexchange.com/questions/1391067/equation-with-absolute-value-and-parameter
