Lesson How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 4

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How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 4


Problem 1

Solve absolute value equation   |x + |3x - 2|| = 2.

Solution 

Starting equation is

    |x + |3x - 2|| = 2.    (1)


It means that 

    either  x + |3x - 2| =  2    (2)

    or      x + |3x - 2| = -2.   (3)


Next consider equations (2) and (3) separately.


       Equation (2)


Equation (2)  is equivalent to

    |3x-2| = 2-x.    (4)



In the domain 3x-2 >= 0,  equation (4) is equivalent to

    3x-2 = 2-x,  3x+x = 2+2,  4x = 4,  x= 4/4 = 1.

    For this value of x,  the expression 3x-2 = 3*1-2 = 3-2 = 1 is positive,
    so, the premise 3x-2 >= 0  is valid;  hence,  x= 1 is a valid solution to equation (4).



In the domain 3x-2 < 0,  equation (4) is equivalent to

    3x-2 = -(2-x),  3x-2 = -2+x,  3x - x = -2 + 2,  2x= 0,  x= 0.

    For this value of x,  the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative,
    so, the premise 3x-2 < 0  is valid;  hence,  x= 0 is a valid solution to equation (4).



       Equation (3)


Equation (3)  is equivalent to

    |3x-2| = -2-x.    (5)



In the domain 3x-2 >= 0,  equation (5) is equivalent to

    3x-2 = -2-x,  3x+x = -2+2,  4x = 0,  x= 0.

    For this value of x,  the expression 3x-2 = 3*0-2 = 0-2 = -2 is negative,
    so, the premise 3x-2 >= 0  is NOT valid;  hence,  x= 0 is NOT a valid solution to equation (5).



In the domain 3x-2 < 0,  equation (5) is equivalent to

    3x-2 = -(-2-x),  3x-2 = 2+x,  3x - x = 2 + 2,  2x= 4,  x= 4/2 = 2.

    For this value of x,  the expression 3x-2 = 3*2-2 = 6-2 = 4 is positive,
    so, the premise 3x-2 < 0  is NOT valid;  hence,  x= 2 is NOT a valid solution to equation (5).


ANSWER.  After this analysis, we see that the only solutions for the given equation (1)  

         are x= 0  and  x= 1.

Notice that this solution follows the strict logic at every step, with the analysis of possible
domains, so it provides the true roots of the original equation without the necessity to check them at the end.

This method of solution and this logic does not create excessive erroneous solutions,
and therefore does not require checking the solutions at the end.

The possible excessive erroneous solutions are rejected (are excluded and filtered out) in the course of analysis.

For visual check, use plotting tool www.desmos/calculator (free of charge for common use)
https://www.desmos.com/calculator/siipdt7pnk


Problem 2

Solve absolute value equation   |3x - |2x + 1|| = 4.

Solution 

Starting equation is

    |3x - |2x + 1|| = 4.    (1)


It means that 

    either  3x - |2x + 1| =  4    (2)

    or      3x - |2x + 1| = -4.   (3)


Next consider equations (2) and (3) separately.


       Equation (2)


Equation (2)  is equivalent to

    |2x+1| = 3x-4.    (4)



In the domain 2x+1 >= 0,  equation (4) is equivalent to

    2x+1 = 3x-4,  1+4 = 3x - 2x,  5 = x,  x= 5.

    For this value of x,  the expression 2x+1 = 2*5+1 = 11 is positive,
    so, the premise 3x-2 >= 0  is valid;  hence,  x= 5 is a valid solution to equation (4).



In the domain 2x+1 < 0,  equation (4) is equivalent to

    2x+1 = -(3x-4),  2x+1 = -3x+4,  2x+3x = 4 - 1,  5x= 3,  x= 3/5.

    For this value of x,  the expression 2x+1 = 3*(3/5)+1 = 9/5+1 is positive,
    so, the premise 2x+1 < 0  is NOT valid;  hence,  x= 3/5 is NOT a valid solution to equation (4).



       Equation (3)


Equation (3)  is equivalent to

    |2x+1| = 3x+4.    (5)



In the domain 2x+1 >= 0,  equation (5) is equivalent to

    2x+1 = 3x+4,  1-4 = 3x-2x,  -3 = x,  x= -3.

    For this value of x,  the expression 2x+1 = 2*(-3)+1 = -6+1 = -5 is negative,
    so, the premise 2x+1 >= 0  is NOT valid;  hence,  x= -3 is NOT a valid solution to equation (5).



In the domain 2x+1 < 0,  equation (5) is equivalent to

    2x+1 = -(3x+4),  2x+1 = -3x-4,  2x + 3x = -4 - 1,  5x= -5,  x= -5/5 = -1.

    For this value of x,  the expression 2x+1 = 2*(-1)+1 = -2+1 = -1 is negative,
    so, the premise 2x+1 < 0  is valid;  hence,  x= -1 is a valid solution to equation (5).


ANSWER.  After this analysis, we see that the only solutions for the given equation (1)  

         are x= -1  and  x= 5.

Notice that this solution follows the strict logic at every step, with the analysis of possible
domains, so it provides the true roots of the original equation without the necessity to check them at the end.

This method of solution and this logic do not create excessive erroneous solutions,
and therefore do not require checking the solutions at the end.

The possible excessive erroneous solutions are rejected (are excluded and filtered out) in the course of analysis.

For visual check, use plotting tool www.desmos/calculator (free of charge for common use)
https://www.desmos.com/calculator/ajbgvspnhp


My other lessons on  Absolute Value equations  in this site are
    - 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

See also the lesson  OVERVIEW of lessons on Absolute Value equations  under the current topic in this site.

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

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