Lesson HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2

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How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 2


In this lesson you will see how to solve equations containing Linear Terms under the Absolute Value sign.
We consider, one after the other, the solution procedure for the following examples:

a)  0.5x-2.5 = -abs%283x-6%29%2B2;
b)  -3x%2B2     = -abs%283x-6%29%2B2;
c)  -3x%2B8     = -abs%283x-6%29%2B2;
d)  0.5x%2B1    = -abs%283x-6%29%2B2;
e)  6x-10      = -abs%283x-6%29%2B2;
f)  3.5           = -abs%283x-6%29%2B2.

The starting lesson in this series is
    How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 1
where simple cases are considered.

This lesson has continuations in the lessons
    How to solve equations containing Linear Terms under the Absolute Value sign. Lesson 3
where more complicated cases are considered.

The goal is to get you acquainted with the different typical situations you can meet solving this kind of problems.
When you learn how to solve these equations, you will easily proceed in many other cases.


Problem 1
Solve absolute value equation  a)  0.5x-2.5 = -abs%283x-6%29%2B2       (1);
Solve absolute value equation  b)  -3x%2B2     = -abs%283x-6%29%2B2       (2);
Solve absolute value equation  c)  -3x%2B8     = -abs%283x-6%29%2B2       (3);
Solve absolute value equation  d)  0.5x%2B1    = -abs%283x-6%29%2B2       (4);
Solve absolute value equation  e)  6x-10      = -abs%283x-6%29%2B2       (5);
Solve absolute value equation  f)  3.5           = -abs%283x-6%29%2B2       (6).

Solution to  1a)

By the definition

    abs%283x-6%29 =    3x-6,   if  x >= 2,  and                                                   
    abs%283x-6%29 =  -%283x-6%29,  if  x < 2.

Hence, we should try to solve the equation

0.5x-2.5 = -%283x-6%29%2B2  over the range  x >= 2,       (7)

and the equation

0.5x-2.5 = -%28-%283x-6%29%29%2B2  over the range  x < 2.     (8)



Figure 1a. To the solution  0.5x-2.5 = -abs%283x-6%29%2B2

Simplifying the equation  (7)  you get  3.5x = 10.5,  and the solution is  x = 3.  It is in the range  x >= 2  and satisfies all the requirements.
Simplifying the equation  (8)  you get  -2.5x = -1.5,  and the solution is  x = 3%2F5.  It is in the range  x < 2  and satisfies all the requirements.
Thus,  the equation  (1)  has two solutions:  x = 2  and  x = 3%2F5.

In  Figure 1a  the green and the black straight lines represent the plot of the function  y = -abs%283x-6%29.  The red line represents the linear function y = 0.5x-2.5.
The intersection points represent the two solutions we just found.


Solution to  1b)

Again, by the definition

    abs%283x-6%29 =    3x-6,   if  x >= 2,  and                                                   
    abs%283x-6%29 =  -%283x-6%29,  if  x < 2.

Hence, we should try to solve the equation

-3x%2B2 = -%283x-6%29%2B2  over the range  x >= 2,         (9)

and the equation

-3x%2B2 = -%28-%283x-6%29%29%2B2  over the range  x < 2.     (10)



Figure 1b. To the solution  -3x%2B2 = -abs%283x-6%29%2B2

Simplifying the equation  (9)  you get  2 = 8,  which means that the equation  (9)  has no solution.
Simplifying the equation  (10)  you get  -6x = -6,  and the solution is  x = 1.  It is in the range  x < 2  and satisfies all the requirements.
Thus,  the equation  (2)  has the unique solution  x = 1.

In  Figure 1b  the green and the black straight lines represent the plot of the function  y = -abs%283x-6%29.  The red line represents the linear function y = -3x%2B2.
The intersection point represents the unique solution we just found.


Solution to  1c)

Again, by the definition

    abs%283x-6%29 =    3x-6,   if  x >= 2,  and                                                   
    abs%283x-6%29 =  -%283x-6%29,  if  x < 2.

Hence, we should try to solve the equation

-3x%2B8 = -%283x-6%29%2B2  over the range  x >= 2,        (11)

and the equation

-3x%2B8 = -%28-%283x-6%29%29%2B2  over the range  x < 2.     (12)



Figure 1c. To the solution  -3x%2B8 = -abs%283x-6%29%2B2

The equation  (11)  is the identity, it is satisfied at any real number  x  of the ramge  x >= 2.
Simplifying the equation  (12)  you get  -6x = -12.  The root is  x = 2,  but it is out of the range  x < 2  and does not satisfy all the requirements.
Thus,  the equation  (3)  has infinitely many solutions.  Every real number of the range x >= 2 is the solution.

In  Figure 1c  the green and the black straight lines represent the plot of the function  y = -abs%283x-6%29.  The red line represents the linear function y = -3x%2B8.
The read straight line coincides with the black straight line at  x>= 2.


Solution to  1d)

This time, we should try to solve the equation                      

0.5x%2B1 = -%283x-6%29%2B2  over the range  x >= 2,       (13)

and the equation

0.5x%2B1 = -%28-%283x-6%29%29%2B2  over the range  x < 0.     (14)



Figure 1d. To the solution  0.5x%2B1 = -abs%283x-6%29%2B2

Simplifying the equation  (13)  you get  4.5x = 9.  The root is x = 2.  This root is in the range  x >= 2  and satisfies all the requirements.
Simplifying the equation  (14)  you get  -2.5x = -5.  It has the root  x = 2,  but this root is out of the range  x < 2  and does not satisfy all the requirements.
Thus,  the equation  (4)  has the unique solution  x = 2.

In  Figure 1d  the green and the black straight lines represent the plot of the function  y = -abs%283x-6%29%2B2.  The red line represents the linear function y = 0.5x%2B1.
The only intersection point represents the unique solution  x = 2  we just found.


Solution to  1e)

This time, we should try to solve the equation                      

6x-10 = -%283x-6%29%2B2  over the range  x >= 2,       (15)

and the equation

6x-10 = -%28-%283x-6%29%29%2B2  over the range  x < 0.     (16)



Figure 1e. To the solution  6x-10 = -abs%283x-6%29%2B2

Simplifying the equation  (13)  you get  9x = 18.  The root is x = 2.  This root is in the range  x >= 2  and satisfies all the requirements.
Simplifying the equation  (14)  you get  3x = 6.  It has the root  x = 2,  but this root is out of the range  x < 2  and does not satisfy all the requirements.
Thus,  the equation  (5)  has the unique solution  x = 2.

In  Figure 1e  the green and the black straight lines represent the plot of the function  y = -abs%283x-6%29%2B2.  The red line represents the linear function y = 6x-10.
The only intersection point represents the unique solution  x = 2  we just found.


Solution to  1f)

This time, we should try to solve the equation                      

3.5 = -%283x-6%29%2B2  over the range  x >= 2       (17),

and the equation

3.5 = -%28-%283x-6%29%29%2B2  over the range  x < 2     (18).



Figure 1f. To the solution  3.5 = -abs%283x-6%29%2B2

Simplifying the equation  (17)  you get  3x = 4.5.  It has the root  x = 1.5,  but this root is out of the range  x >= 2  and does not satisfy all the requirements.
Simplifying the equation  (18)  you get  -3x = -7.5.  It has the root  x = 2.5,  but this root is out of the range  x < 2  and does not satisfy all the requirements.
Thus,  the equation  (6)  has not the solution.

In  Figure 1f  the green and the black straight lines represent the plot of the function  y = abs%28x%29.  The red line represents the linear function y = 3.5.
There is no intersection point of the plot  y = -abs%283x-6%29%2B2  and the plot  y = 3.5.


Below the plots of the  Figure 1a - Figure 1f  are presented again for your convenience.



Figure 1a. To the solution  0.5x-2.5 = -abs%283x-6%29%2B2    
                  (two roots)


Figure 1b. To the solution  -3x%2B2 = -abs%283x-6%29%2B2    
              (the unique root)


Figure 1c. To the solution  -3x%2B8 = -abs%283x-6%29%2B2
        (infinitely many roots)



Figure 1d. To the solution  0.5x%2B1 = -abs%283x-6%29%2B2        
                  (one root)


Figure 1e. To the solution  6x-10 = -abs%283x-6%29%2B2    
                  (one root)


Figure 1f. To the solution  3.5 = -abs%283x-6%29%2B2
                  (no roots)


Note that the strategy of the solution is the same in all examples.  It is to break up the entire set of real numbers into sub-domains (ranges)
where the absolute value of a linear term is a linear function,  and then to solve the corresponding linear equations in each sub-domain (range).


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 3
    - HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 4

    - 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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