Exercises for Absolute Value Solutions
A) Solve each of the following. To solve means to find and write the solution set. Write each
solution set in as many ways as you can. Graph each of the inequalities.
1) 3 < 2x – 5 < 9
Analysis:
3  2x  5  9
8  2x  14
4x7
The solution set is the interval  4, 7  = x 4  x  7
2) – 4 < 3x + 7 < 4
Analysis:
4  3x  7  4
11  3x  3
 11  x  1
3

 

The solution set is the interval  11 , 1  x  11  x  1
3
3
3) – 1 < x +1 < 1
Analysis:
1  x  1  1
2  x  0
The solution set is the interval  2,0  x 2  x  0
4) – 5 < 2x – 3 < 5
Analysis:
5  2x  3  5
2  2x  8
1  x  4
The solution set is the interval  1, 4   x 1  x  4
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B) Solve each of the following without doing any computation. Explain why this is the solution
set.
1) |2x – 7| < – 8
Analysis:
The solution set for |2x – 7| < – 8 is the empty set  because the absolute value cannot be negative.
2) |5x + 2| < 0
Analysis:
The solution set for |5x + 2| < 0 is the empty set  because the absolute value cannot be negative.
3) |x + 3| < – 2
Analysis:
The solution set for |x + 3| > –2 is the empty set R because the absolute value is non-negative.
4) |– 9x – 6| < –1
Analysis:
The solution set for |– 9x – 6| > – 1 is the empty set R because the absolute value is non-negative.
C) Write each of the following equations as an equivalent linear equation. Explain why the two
equations are equivalent.
1) |2x – 7| = 0
Analysis:
The equation |2x – 7| = 0 is equivalent to 2x – 7 = 0 because |a| = 0 if and only if a = 0.
2) |5x + 2| = 0
Analysis:
The equation |2x – 7| = 0 is equivalent to 2x – 7 = 0 because |a| = 0 if and only if a = 0.
3) |x + 3| = 0
Analysis:
The equation |2x – 7| = 0 is equivalent to 2x – 7 = 0 because |a| = 0 if and only if a = 0.
4) |– 9x – 6| = 0
Analysis:
The equation |2x – 7| = 0 is equivalent to 2x – 7 = 0 because |a| = 0 if and only if a = 0.
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D) Solve each of following inequalities. Explain your work.
1) |3x – 5| > 0
Analysis:
    

5,  x x  5
The solution set for |3x – 5| > 0 is , 5
3
3
3
Because |a| > 0 for all real numbers a except a = 0.
2) |2x + 7| > 0
Analysis:

 
 
The solution set for |2x + 7| > 0 is ,  7
 7 ,  x x   7
2
2
2
Because |a| > 0 for all real numbers a except a = 0.
3) |3x – 1| > 0
Analysis:
    


1 ,  x x  1
The solution set for |3x – 1| > 0 is , 1
3
3
3
Because |a| > 0 for all real numbers a except a = 0.
4) |10x – 17| > 0
Analysis:

   
17 ,   x x  17
The solution set for |10x – 17| > 0 is , 17
10
10
10
Because |a| > 0 for all real numbers a except a = 0.
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
E) Write each of the following inequalities as an equivalent compact compound inequality
without absolute value. Solve the compact compound inequality computationally. Write the
solution set in interval notation and set builder notation. Sketch the graph of the compact
compound inequality. Write the corresponding boundary equation. Use set builder notation to
write the solution set of the boundary equation. Sketch the graph of the boundary equation.
Write the corresponding “greater than” inequality. Use both interval notation and set builder
notation to write the solution set of the “greater than” inequality. Sketch the graph of the
“greater than” inequality. Now sketch the three graphs on the same number line, making sure to
properly label each of the individual graphs. What is the union of the three solution sets? What
is the intersection of any pair of these solution sets?
1) |3x – 5| < 4
Analysis:
3x  5  4 is equivalent to
4  3x  5  4
1  3x  9
1x3
3
  

The solution set for 3x  5  4 is the interval 1 , 3  x 1  x  3
3
3
 
The boundary equation is 3x  5  4 has the solution set 1 , 3
3
The greater than inequality is 3x  5  4 whose solution set is everything else.

The solution set for 3x  5  4 is , 1
3

 3,   


x x  1 or x  3
3
The three graphs may be combined to more completely illustrate The Law of Trichotomy.
The union of the three solution sets is R.
The intersection of any two of the solution set is the empty set .
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2) |5x – 7| < 1
Analysis:
5x  7  1 is equivalent to
1  5x  7  1
6  3x  10
2  x  10
3
  
The solution set for 5x  7  1 is the interval 2, 10  x 2  x  10
3
3

 
The boundary equation is 5x  7  1 has the solution set 2, 10
3
The greater than inequality is 5x  7  1 whose solution set is everything else.
The solution set for 5x  7  1 is  , 2 
103 ,   x x  2 or x  103 
The three graphs may be combined to more completely illustrate The Law of Trichotomy.
The union of the three solution sets is R.
The intersection of any two of the solution set is the empty set .
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3) |5x + 1| < 9
Analysis:
5x  1  9 is equivalent to
9  5x  1  9
10  5x  8
2  x  8
5
  
The solution set for 5x  1  1 is the interval 2, 8  x 2  x  8
5
5
 

The boundary equation is 5x  1  9 has the solution set 2, 8
5
The greater than inequality is 5x  1  9 whose solution set is everything else.
The solution set for 5x  1  9 is  , 2 
 85 ,   x x  2 or x  85
The three graphs may be combined to more completely illustrate The Law of Trichotomy.
The union of the three solution sets is R.
The intersection of any two of the solution set is the empty set .
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4) |x – 17| < 2
Analysis:
x  17  2 is equivalent to
2  x  17  2
15  x  19
15  x  19
The solution set for x  17  2 is the interval 15,19   x 15  x  19
The boundary equation is x  17  2 has the solution set 15,19
The greater than inequality is x  17  2 whose solution set is everything else.
The solution set for x  17  2 is  ,15
19,    x x  15 or x  19
The three graphs may be combined to more completely illustrate The Law of Trichotomy.
The union of the three solution sets is R.
The intersection of any two of the solution set is the empty set .
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F) Solve each of the inequalities by computationally solving “the easy one” and then deducing
solution sets for the given inequality. Sketch the graph of the given inequality.
1) |3x – 5|  4
Analysis:
The desired solution set is the union of the solution set for |3x – 5| < 4 and the solution set for the
boundary equation |3x – 5| = 4.
The inequality |3x – 5| < 4 is equivalent to
4  3x  5  4
1  3x  9
1x3
3
The solution set for | 3x  5 | 4 is the interval 1 ,3 .
3
 
 
The solution set for the boundary equation is the set 1 ,3 of endpoints of that interval.
3
  13 ,3   13 ,3
The desired solution set is the union 1 ,3
3
The graph of |3x – 5|  4 is
2) |5x – 7|  1
Analysis:
The desired solution set is the union of the solution set for |5x – 7| > 1 and the solution set for the
boundary equation |5x – 7| = 1.
Begin by solving the easy inequality |5x – 7| < 1which is equivalent to
1  5x  7  1
6  5x  8
6x8
5
5
 
The solution set for 5x  7  1 is the interval 6 , 8 .
5 5
 
The solution set for the boundary equation is the set 6 , 8 of endpoints of that interval.
5 5
8,
The solution set for the inequality 5x  7  1 is everything else: , 6
5
5
   
The desired solution set is the union   , 6   8 ,    6 , 8    , 6   8 ,  
5
5
5   5

 5 5
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The graph of |5x – 7|  1 is
3) |5x + 1|  9
Analysis:
The desired solution set is the union of the solution set for |5x + 1| < 9 and the solution set for the
boundary equation |5x + 1| = 9.
The inequality |5x + 1| < 9 is equivalent to
9  5x  1  9
10  5x  8
2  x  8
5
The solution set for | 5x  1 | 9 is the interval 2, 8 .
5
 
The solution set for the boundary equation is the set 2, 8 of endpoints of that interval.
5
The desired solution set is the union  2, 8  2, 8   2, 8 
5
5 
5 
The graph of |5x +1|  9 is
4) |x – 17|  2
Analysis:
The desired solution set is the union of the solution set for |x – 17| > 2 and the solution set for the
boundary equation |x – 17| = 2.
Begin by solving the easy inequality |x – 17| < 2 which is equivalent to
2  x  17  2
15  x  19
The solution set for x  17  2 is the interval 15,19  .
The solution set for the boundary equation is the set 15,19 of endpoints of that interval.
The solution set for the inequality x  17  2 is everything else:  ,15
The desired solution set is the union
19,  
  ,15 19,    15,19   ,15 19,  
The graph of |5x – 7|  1 is
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