Question 1094264
.
|5x -4| = |x + 17|           (1)



        As you know, the absolute value functions have different expressions in different parts of the number line.

        The strategy of solving such problems is to divide the number line in segments/intervals, 

        where the absolute value functions can be written in more simple way, and then solve the equations over each such interval.


        Below find an implementation of this strategy.


<pre>
There are 2 critical points: x = {{{4/5}}} and x = -17,  and 3 intervals:

(-infinity,-17],  (-17,{{{4/5}}})  and  [{{{4/5}}},infinity).


1)  Interval (-infinity,-17].

    In this interval,  5x-4 < 0;    therefore, |5x-4| = -(5x-4).
                       x + 17 <=0;  therefore, |x+17| = -(x+17).

    Therefore, the original equation (1) takes the form

        -(5x-4) = -(x+17)   over the domain  x <= -17.    (2)

     Simplify:

        -5x + 4 = -x - 17  ====>  4 + 17 = -x + 5x  ====>  4x = 21  ====>  x = {{{21/4}}}. 

      Now notice that this value {{{21/4}}}  <U>does not belong</U>  to the domain x <= -17.  So, the equation (2) HAS NO solution/solutions over this domain.


      Thus this case is completed.



2)  Interval (-17,{{{4/5}}}).

    In this interval,  5x-4 < 0;    therefore, |5x-4| = -(5x-4).
                       x + 17 > 0;  therefore, |x+17| =   x+17.

    Therefore, the original equation (1) takes the form

        -(5x-4) = x+17   over the domain   -17 < x < {{{4/5}}}.    (3)

     Simplify:

        -5x + 4 = x + 17  ====>  4 - 17 = x + 5x  ====>  6x = -13  ====>  x = {{{-13/6}}}. 

      This value {{{-13/6}}}  <U>does belong</U>  to the domain -17 < x < {{{4/5}}}.  
      So, the equation (3) HAS the solution {{-13/6}}} in this domain.
      Hence, it is the solution to the original equation (1), too.

      Thus this case is completed.



3)  Interval [{{{4/5}}},infinity).

    In this interval,  5x-4 > 0;    therefore, |5x-4| =   5x-4.
                       x + 17 > 0;  therefore, |x+17| =   x+17.

    Therefore, the original equation (1) takes the form

         5x-4 = x+17   over the domain  x > {{{4/5}}}.    (4)

     Simplify:

         5x-4 = x+17  ====>  5x - x = 17+4 = 21  ====>  4x = 21  ====>  x = {{{21/4}}}. 

      This value {{{21/4}}}  <U>does belong</U>  to the domain  x > {{{4/5}}}.  
      So, the equation (4) HAS the solution {{21/4}}} in this domain.
      Hence, it is the solution to the original equation (1), too.

      Thus this case is completed, too.


<U>Answer</U>.  The original equation (1) has two solutions:  x = {{{-13/6}}}   and  x = {{{21/4}}}.
</pre>

Solved.



{{{graph( 330, 330, -25.5, 10.5, -5.5, 30.5,
          abs(5x-4), abs(x+17)
)}}}


Plot y = |5x-4| (red) and y = |x+17| (green)



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On solving &nbsp;<B>Absolute Value equations</B>&nbsp; see the lessons in this site

&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.





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