SOLUTION: HOW do I do this?: Solve each absolute value inequality and show it's solution set: a)|x|&#8804;3 b)|3&#8722;x|<5 c)|4g&#8722;5|>17 d)|4x+1|<21

Algebra ->  Finance -> SOLUTION: HOW do I do this?: Solve each absolute value inequality and show it's solution set: a)|x|&#8804;3 b)|3&#8722;x|<5 c)|4g&#8722;5|>17 d)|4x+1|<21      Log On


   



Question 1116225: HOW do I do this?:
Solve each absolute value inequality and show it's solution set:
a)|x|≤3
b)|3−x|<5
c)|4g−5|>17
d)|4x+1|<21

Found 3 solutions by stanbon, greenestamps, ikleyn:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
HOW do I do this?:
Solve each absolute value inequality and show it's solution set:
a)|x|≤3
-3<= x <=3
====================
b)|3−x|<5
-5 < 3-x < 5
-9 < -x > 2
-2 < x < 9
====================
c)|4g−5|>17
4g-5 < -17 or 4g-5 > 17
g < -3 or g > 11/2
------------------------------
d)|4x+1|<21
-21 < 4x+1 < 21
-22 < 4x < 20
-11/2 < x < 5
------------------
Cheers,
Stan H.
------------------

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(1) The inequality

abs%28x-a%29%3C=b

where b is a non-negative number, means the value of x-a is between -b and b:

-b+%3C=+x-a+%3C=+b

As an example, suppose the inequality is

abs%28x-3%29+%3C=+7

We can find the solution set using the pattern shown above:

-7+%3C=+x-3+%3C=+7
-4+%3C=+x+%3C=+10

The solution set in interval notation is [-4,10]

(2) Here is another way of interpreting the same inequality

This interpretation is helpful when the inequalities get more complicated.

The inequality

abs%28x-a%29%3C=b

means the difference between x and a is less than or equal to b. On a number line, that means x can be any number whose distance from a is at most b.

So the same example as above,

abs%28x-3%29+%3C=+7

has as its solution set all the numbers whose distance from 3 is at most 7.

This is easy to see on a number line; 7 to the left of 3 is -4, and 7 to the right is 10, so the solution set is all values of x from -4 to 10.

Note that for the inequality abs%28x-3%29+%3E+7 the solution set is all values of x that are OUTSIDE the interval from -4 to 10. So the solution set for that inequality would be (-infinity, -4) union (10,infinity).

Solving that kind of inequality with the first method requires solving two equations separately; with the second method you simply have a solution set that is all the values that are NOT between -4 and 10 on the number line.

You can use either of these interpretations to solve each of your examples. If you really want to learn about solving inequalities, you should solve each of your examples by BOTH methods.

Here are some hints for helping you with your examples:
(a) abs%28x%29 is the same as abs%28x-0%29
(b) abs%283-x%29 is the same as abs%28x-3%29
(c) You can divide both sides of an inequality by a positive constant. So abs%284g-5%29%3E17 is equivalent to abs%28g-5%2F4%29%3E17%2F4. Note also in this problem the solution set is the set of values for which the absolute value of the expression is GREATER THAN some number. So the solution set will be the set of all numbers OUTSIDE of some interval -- that is, it will be the union of two intervals.
(d) Again divide both sides by 4 to get abs%28x%2B1%2F4%29%3C21%2F4

Answer by ikleyn(52785) About Me  (Show Source):
You can put this solution on YOUR website!
.
Since the post by @stanbon contained errors,  I copied,  pasted,  corrected it and presented in more readable format:

HOW do I do this?:
Solve each absolute value inequality and show it's solution set:

a) |x| <= 3

   -3<= x <=3

====================

b) |3-x| < 5

   -5 < 3-x < 5
   -8 < -x < 2    
   -2 < x < 8    <<<---=== those "x", that are remoted less than 5 units from the point  "3" in the number line.

====================

c) |4g-5| > 17

   4g-5 < -17  or  4g-5 > 17
   g < -3 or g > 11/2           <<<---=== the solution is the union of two sets

====================

d) |4x+1| < 21 

   -21 < 4x+1 < 21
   -22 < 4x < 20
   -11/2 < x < 5

-------------
For those who learn the solution of absolute value equations for the first time in their life,  there is a group of lessons
    - Absolute Value equations
    - HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1
    - HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2
    - HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3
in this site.

Read them attentively and become an expert in this area.   Start from the very first lesson of the list.


Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic
"Solving Absolute values equations".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.