Question 1139380
.


            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.



<pre>
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/(-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.




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


<U>ANSWER</U>.  The given inequality HAS NO SOLUTIONS.
</pre>


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



<pre>
{{{graph( 330, 330, -5, 5, -5, 10,
          abs(x+3) + abs(x-3), 5.9
)}}}


Plot y = | x+3 | + | x-3 | (red)  and  y = 6  (green).
</pre>

Solved.


--------------


To see many other similar solved problems, look into the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/-Absolute-Value-equations.lesson>Absolute Value equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-2.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-3.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-2.lesson>HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/absolute-value/Review-of-lessons-on-Absolute-Value-equations.lesson>OVERVIEW of lessons on Absolute Value equations</A> 


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

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


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



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.