SOLUTION: Solve and graph the solution sets... |h+3| + |h-3| < 6

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Question 1139380: Solve and graph the solution sets...
|h+3| + |h-3| < 6
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


abs%28h%2B3%29 = -%28h%2B3%29+=+-h-3 if h < -3;
abs%28h%2B3%29 = %28h%2B3%29 if h > -3

abs%28h-3%29 = -%28h-3%29+=+-h%2B3 if h < 3;
abs%28h-3%29 = %28h-3%29 if h > 3

So for h < -3,

abs%28h%2B3%29%2Babs%28h-3%29+=+%28-h-3%29%2B%28-h%2B3%29+=+-2h;

for -3 < h < 3,
abs%28h%2B3%29%2Babs%28h-3%29+=+%28h%2B3%29%2B%28-h%2B3%29+=+6;

and for h > 3,
abs%28h%2B3%29%2Babs%28h-3%29+=+%28h%2B3%29%2B%28h-3%29+=+2h

For values of x less than -3, the value of the expression is -2h, which means the value of the expression is always greater than 6.

For values of x between -3 and 3, the value of the expression is always 6.

For values of x greater than 3, the value of the expression is 2h, which means the value of the expression is again always greater than 6.

So the value of the expression is equal to or greater than 6 for ALL values of x.

And therefore the solution set for the given inequality is the empty set.

A graph....

graph%28400%2C400%2C-5%2C5%2C-5%2C10%2Cabs%28x%2B3%29%2Babs%28x-3%29%2C6%29

Answer by ikleyn(52831) About Me  (Show Source):
You can put this solution on YOUR website!
.

            This inequality has two linear functions under the absolute value sign each.

            This "absolute value" sign transform a linear function into non-linear, which is not so simple to solve.

            Therefore, the solution strategy is to divide the entire number line into separate intervals / segments
            in a way that at each interval/segment an absolute value function is LINEAR.

            Then the solution is doable and simple.

            Below is how I implement this idea. 


In this case we have two critical points, x= -3  and x= 3, that divide the entire number line in 3 non-intersecting intervals/segments


    1)  h < -3;    2)  -3 <= h <= 3;   and   3)  h > 3.


Let's analyze each interval separately.



1)  If h < -3,  then  | h+3| = -(h+3)  and  | h-3 | = -(h-3).


                therefore, the original inequality takes the form

                    -(h+3) + (-(h-3) < 6.

                Simplify and solve it step by step

                     -h - 3 - h + 3 < 6

                     -2h < 6

                       h > 6%2F%28-2%29 = -3.

               So, we started from  h < -3 and obtained h > -3.
               It means that in the interval h < -3 the original inequality HAS NO solutions.




2)  If -3 <= h <= 3,  then  | h+3| = h+3  and  | h-3 | = -(h-3).


                therefore, the original inequality takes the form

                    h+3 + (-(h-3) < 6.

                Simplify and solve it step by step

                      h + 3 - h + 3 < 6

                      6 < 6


               It is SELF-CONTRADICTING inequality, and it HAS NO SOLUTIONS.
               It means that in the interval -3 <= h <= 3 the original inequality HAS NO solutions.



2)  If  h > 3,  then  | h+3| = h+3  and  | h-3 | = h-3.


                therefore, the original inequality takes the form

                    h+3 + h-3 < 6.

                Simplify and solve it step by step

                      h + h  < 6

                      2h < 6

                      h < 6/2 = 3/


               So, we started from  h > 3 and obtained h < 3.
               It means that in the interval h > 3 the original inequality HAS NO solutions.


Thus, after completing analyses of all 3 cases/intervals we come to this conclusion


ANSWER.  The given inequality HAS NO SOLUTIONS.


See the plot below, which visually shows that the original inequality is NEVER true. 





Plot y = | x+3 | + | x-3 | (red)  and  y = 6  (green).

Solved.

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To see many other similar solved problems, look into the 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
    - HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1
    - HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2
    - OVERVIEW of lessons on Absolute Value equations

Read them attentively and become an expert in this area.


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.