Question 1139420
.


            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= {{{-1/2}}}  and x= {{{4/3}}}, where the functions change their linear behavior.

These points divide the entire number line in 3 non-intersecting intervals/segments


    1)  x < {{{-1/2}}};    2)  {{{-1/2}}} <= x <= {{{4/3}}};   and   3)  x > {{{4/3}}}.


Let's analyze each interval separately.



1)  If x < {{{-1/2}}},  then  | 2x+1 | = -(2x+1)  and  | 4-3x | = 4-3x.


                therefore, the original inequality takes the form

                    (-3)*(-(2x+1)) + 2*(4-3x) < 7.

                Simplify and solve it step by step

                     6x + 3 + 8 - 6x < 7

                     11 < 7.


               This inequality is FALSE.
               It means that the interval  x < {{{-1/2}}} is NOT the solution to the original inequality.



2)  If {{{-1/2}}} <= x <= {{{4/3}}},  then  | 2x+1 | = 2x+1  and  | 4-3x | = 4-3x.


                therefore, the original inequality takes the form

                    (-3)*(2x+1) + 2*(4-3x) < 7.

                Simplify and solve it step by step

                     -6x - 3 + 8 - 6x < 7

                     -12x < 2.

                      x   > {{{2/(-12)}}} = {{{-1/6}}}


               It means that in the interval {{{-1/6}}} <= x <= {{{4/3}}} is the partial solution to the original inequality.



2)  If  x > {{{4/3}}},  then  | 2x+1 | = 2x+1  and  | 4-3x | = -(4-3x).


                therefore, the original inequality takes the form

                    (-3)*(2x+1) + 2*(-(4-3x)) < 7.

                Simplify and solve it step by step

                     -6x - 3 - 8 + 6x < 7

                     -11 < 7.


               This inequality is TRUE.
               It means that in the interval x >= {{{4/3}}} is the partial solution to the original inequality.


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


<U>ANSWER</U>.  The given inequality has the solution set  x >= {{{-1/6}}}.
</pre>


See the plot below, which visually confirms the found solution.



<pre>
    {{{graph( 330, 330, -3, 3, -12, 12,
          2*(abs(4-3x)) - 3*(abs(2x+1)), 7
)}}}


    Plot y = 2*|4-3x| - 3*|2x+1| (red)  and  y = 7  (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.