Solving Absolute Value Inequalities
As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at ( −x,0 ) ( −x,0 ) has an absolute value of x, x, as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero.
An absolute value inequality is an equation of the form
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|A|<B,|A|≤B,| A |>B,or | A |≥B, |A|<B,|A|≤B,|A|>B,or |A|≥B,
Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all x x -values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values.
There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph.
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Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between x x and 600 is less than or equal to 200. We represent the distance between x x and 600 as | x−600 |, | x−600 |, and therefore, | x−600 |≤200 | x−600 |≤200 or
−200≤x−600≤200 −200+600≤x−600+600≤200+600 400≤x≤800 −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.
The distance from x x to 5 can be represented using an absolute value symbol, | x−5 |. | x−5 |. Write the values of x x that satisfy the condition as an absolute value inequality.
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| x−5 |≤4 | 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 x−5≤4 and x−5≥−4 x≤9 x≥1
If the solution set is x≤9 x≤9 and x≥1, x≥1, then the solution set is an interval including all real numbers between and including 1 and 9.
So | x−5 |≤4 | x−5 |≤4 is equivalent to [ 1,9 ] [ 1,9 ] in interval notation.
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