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−200 x 600 200 −200 + 600 x 600 + 600 200 + 600 400 x 800

This means our returns would be between $400 and $800.

To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently.

Absolute value inequalities

For an algebraic expression X, and k > 0 , an absolute value inequality is an inequality of the form

| X | < k  is equivalent to  k < X < k | X | > k  is equivalent to  X < k  or  X > k

These statements also apply to | X | k and | X | k .

Determining a number within a prescribed distance

Describe all values x within a distance of 4 from the number 5.

We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as in [link] , to represent the condition to be satisfied.

A number line with one tick mark in the center labeled: 5.  The tick marks on either side of the center one are not marked.  Arrows extend from the center tick mark to the outer tick marks, both are labeled 4.

The distance from x to 5 can be represented using an absolute value symbol, | x 5 | . Write the values of x that satisfy the condition as an absolute value inequality.

| x 5 | 4

We need to write two inequalities as there are always two solutions to an absolute value equation.

x 5 4 and x 5 4 x 9 x 1

If the solution set is x 9 and x 1 , then the solution set is an interval including all real numbers between and including 1 and 9.

So | x 5 | 4 is equivalent to [ 1 , 9 ] in interval notation.

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Describe all x- values within a distance of 3 from the number 2.

| x −2 | 3

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Solving an absolute value inequality

Solve | x 1 | 3 .

| x 1 | 3 −3 x 1 3 −2 x 4 [ −2 , 4 ]
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Using a graphical approach to solve absolute value inequalities

Given the equation y = 1 2 | 4 x 5 | + 3 , determine the x -values for which the y -values are negative.

We are trying to determine where y < 0 , which is when 1 2 | 4 x 5 | + 3 < 0. We begin by isolating the absolute value.

1 2 | 4 x 5 | < 3 Multiply both sides by –2, and reverse the inequality . | 4 x 5 | > 6

Next, we solve for the equality | 4 x 5 | = 6.

4 x 5 = 6 4 x 5 = 6 4 x = 11 or 4 x = 1 x = 11 4 x = 1 4

Now, we can examine the graph to observe where the y- values are negative. We observe where the branches are below the x- axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at x = 1 4 and x = 11 4 , and that the graph opens downward. See [link] .

A coordinate plan with the x-axis ranging from -5 to 5 and the y-axis ranging from -4 to 4.  The function y = -1/2|4x – 5| + 3 is graphed.  An open circle appears at the point -0.25 and an arrow
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Solve 2 | k 4 | 6.

k 1 or k 7 ; in interval notation, this would be ( , 1 ] [ 7 , ) .

A coordinate plane with the x-axis ranging from -1 to 9 and the y-axis ranging from -3 to 8.  The function y = -2|k  4| + 6 is graphed and everything above the function is shaded in.
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Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities.

Key concepts

  • Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See [link] and [link] .
  • Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See [link] , [link] , [link] , and [link] .
  • Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See [link] and [link] .
  • Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See [link] and [link] .
  • Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See [link] .

Questions & Answers

The sequence is {1,-1,1-1.....} has
amit Reply
circular region of radious
Kainat Reply
how can we solve this problem
Joel Reply
Sin(A+B) = sinBcosA+cosBsinA
Eseka Reply
Prove it
Eseka
Please prove it
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
Arleathia Reply
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
Prince Reply
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
SANDESH Reply
write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
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Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
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Aaron Reply
the polar co-ordinate of the point (-1, -1)
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Practice Key Terms 4

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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