MATH 140 -- Intermediate Algebra -- Exercise Solutions
Section: 2.6

2. Solve |y| = 15
Solution:
From the definition of absolute value we have |y| = y if y ≥ 0 or |y| = - y if y < 0
So the original equation is equivalent to the two equations y = 15 or - y = 15
Multiply both sides of the second equation by -1 to obtain the two simplest equations
y = 15 or y = -15.
The solution set is {15, -15}

4. Solve |6n| = 12.6
Solution:  
From the definition of absolute value we have |6n| = 6n if y ≥ 0 or |6n| = - 6n if 6n < 0
So the original equation is equivalent to the two equations 6n = 12.6 or - 6n = 12.6
Multiply both sides of the second equation by - 1 to obtain 6n = 12.6 or   6n = - 12.6
Divide both sides of the each equation by 2.1.
to obtain the two simplest equations
y = 2.1 or y = - 2.1.
The solution set is {2.1, - 2.1}

6. Solve |6 + 2n| = 4
Solution:
From the definition of absolute value we have
|6 + 2n| = 6 + 2n if 6 + 2n ≥ 0    or |6 + 2n| = - (6 + 2n) if 6 + 2n< 0
So the original equation is equivalent to the two equations 6 + 2n = 4 or - (6 + 2n) = 4
Multiply both sides of the second equation by -1 to obtain 6 + 2n = 4 or (6 + 2n) = - 4
Subtract 6 from both sides of each equation to obtain 2n = - 2   or 2n = -10
Divide both sides of each equation to obtain the two simplest equations n = -1 or n = - 5.
The solution set is {- 1, - 5}

8.  Solve
Solution:
Starting with the definition of absolute value applied to this problem and arguing as in the previous exercises we arrive at the following two equations:
     
Solving these two equations yields:

The solution set is {6, -18}

10.  Solve |x| +1 = 3
Solution:
|x| +1 = 3
|x| = 2
x = 2 or x = - 2
The solution set is {-2, 2}

12. Solve |2x| - 6 = 4
Solution:
|2x| - 6 = 4
|2x| = 10
2x = 10 or 2x = - 10
x = 5  or  x = - 5
The solution set is {- 5, 5}

14. Solve |7z| = 0
Solution:
|7z| = 0 fi and only if 7z = 0 if and only if z = 0
The solution set is {0}

16. Solve |3z - 2| + 8 = 1
Solution:
|3z - 2| + 8 = 1
|3z - 2| = - 7
This is impossible because the absolute value is never negative.
The solution set is

18. Solve |3y + 2| = 0
Solution:
|3y + 2| = 0 if an only if 3y + 2 = 0
3y + 2 = 0
3y = -2
y = - 2/3
The solution set is

22. Solve |9y + 1| = |6y + 4|
Solution:
|9y + 1| = |6y + 4|
9y + 1 = 6y + 4   or    9y + 1 = - (6y + 4)
3y + 1 = 4            or    15y + 1 = - 4
3y = 3                   or    15y = - 5
y = 1                     or     y = - 1/3
The soultion set is

24. Solve |2x - 5| = |2x + 5|
Solution:
|2x - 5| = |2x + 5|
2x - 5 = 2x + 5    or    2x - 5 = - (2x + 5)
- 5 = 5                  or    4x - 5 = - 5
no solutions     or     4x = 0
no solutions     or       x = 0
The solution set is {0}