Lesson Advanced problems on solving absolute value inequalities

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Advanced problems on solving absolute value inequalities


Problem 1

Solve an inequality   |2*|x-3|-7| < 5.

Solution 

    | 2*(|x-3|)-7 | < 5


is equivalent to


    -5 < 2*|x-3| - 7 < 5


is equivalent  (after adding 7 to each of 3 parts of the inequality)


    -5 + 7 < 2*|x-3| < 5 + 7


is the same as


    2 < 2*|x-3| < 12    


is equivalent to   (after dividing all the three terms of the inequality by 2)


    1 < |x-3| < 6.


The solutions to the last inequality are those values of x (those numbers or points in the number line) 
that are remoted from the point x= 3 farther than 1 unit and closer than 6 units.


Obviously, these points (numbers, solutions) are 


    -3 < x < 2   and/or  4 < x < 9.


ANSWER.  The solution set to the original inequality is the union  of two intervals  {-3,2)  and  (4,9):  (-3,2) U (4,9).

Problem 2

Solve an inequality   |3*|x-5|-8| > 5.

Solution 

    | 3*(|x-5|)-8 | > 5


is equivalent to the system of two inequalities


    3*|x-5| - 8 > 5     (a)

OR 

    3*|x-5| - 8 < -5    (b)



Notice this service word "OR" in the formulation of the system.

It means that you should solve each inequality and then the final solution set to the system is the UNION of the two individual solution sets.



Now I will solve inequalities (a) and (b) separately, and will start with inequality (a).


    3*|x-5| - 8 > 5     (a)


is the same as


     3*|x-5| > 5 + 8 = 13 


is equivalent to   (after dividing all the three terms of the inequality by 3)


     |x-5| > 13%2F3.


The solutions to the last inequality are those values of x (those numbers or points in the number line) 
that are remoted from the point x= 5 farther than 13%2F3 units.


Obviously, these points (numbers, solutions) are 


    x < 5 - 13%2F3   and/or  x > 5 + 13%2F3.



Next I will solve inequality (b).


    3*|x-5| - 8 < -5     (b)


is the same as


     3*|x-5| < -5 + 8 = 3 


is equivalent to   (after dividing all the three terms of the inequality by 2)


     |x-5| < 1.


The solutions to the last inequality are those values of x (those numbers or points in the number line) 
that are remoted from the point x= 5 closer than 1 units.


Obviously, these points (numbers, solutions) are  4 < x < 6,  or the interval  (4,6).



After completing solving inequalities (a) and (b) we have the


ANSWER.  The solution set to the original inequality is the union  of three intervals  x < 5 - 13%2F3,  4 < x < 6,  x > 5 + 13%2F3.

Problem 3

Solve an inequality   |h+3| + |h-3| < 6.

Solution

            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 straightforward.

            Below is how I implemented this idea. 


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%2F%28-2%29 = -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


ANSWER.  The given inequality HAS NO SOLUTIONS.


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





Plot y = | x+3 | + | x-3 | (red)  and  y = 6  (green).

Problem 4

Solve an inequality   2|4-3x|-3|2x+1| < 7.

Solution

            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 implemented this idea. 


In this case we have two critical points, x= -1%2F2  and x= 4%2F3, where the functions change their linear behavior.

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


    1)  x < -1%2F2;    2)  -1%2F2 <= x <= 4%2F3;   and   3)  x > 4%2F3.


Let's analyze each interval separately.



1)  If x < -1%2F2,  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%2F2 is NOT the solution to the original inequality.



2)  If -1%2F2 <= x <= 4%2F3,  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%2F%28-12%29 = -1%2F6


               It means that in the interval -1%2F6 <= x <= 4%2F3 is the partial solution to the original inequality.



2)  If  x > 4%2F3,  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%2F3 is the partial solution to the original inequality.


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


ANSWER.  The given inequality has the solution set  x >= -1%2F6.


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


    


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


My other lessons on solving inequalities are
    - Solving simple and simplest linear inequalities
    - Solving absolute value inequalities
    - Solving systems of linear inequalities in one unknown
    - Solving compound inequalities

    - What number is greater? Comparing magnitude of irrational numbers
    - Arithmetic mean and geometric mean inequality
    - Arithmetic mean and geometric mean inequality - Geometric interpretations
    - Harmonic mean
    - Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c)

    - Solving problems on quadratic inequalities
    - Solving inequalities for high degree polynomials factored into a product of linear binomials
    - Solving inequalities for rational functions with numerator and denominator factored into a product of linear binomials
    - Solving inequalities for rational functions with non-zero right side
    - Another way solving inequalities for rational functions with non-zero right side

    - Advanced problems on inequalities
    - Challenging problems on inequalities
    - Solving systems of inequalities in two unknowns graphically in a coordinate plane
    - Solving word problems on inequalities
    - Proving inequalities
    - Math circle level problem on inequalities
    - Math Olympiad level problems on inequalities
    - Entertainment problems on inequalities
under the topic  Inequalities, trichotomy of the section  Algebra-I.

My lessons on domains of functions are
    - Domain of a function which is a quadratic polynomial under the square root operator
    - Domain of a function which is a high degree polynomial under the square root operator
    - Domain of a function which is the square root of a rational function.
under the topic  Functions, Domain  of the section  Algebra-I.

See also  OVERVIEW of lessons on inequalities and domains of functions.

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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