Solve Equations with Absolute Value

Example 1: Solve the equation

Solution to Example 1:

• If | x + 6 | = 7, then
  a) x + 6 = 7
  or
  b) x + 6 = -7
  
  
• Solve equation a)
  x + 6 = 7
  x = 1
  
  
• Solve equation b)
  x + 6 = -7
  x = -13
  
  

Check solutions:

• solution x = 1
  Left Side of Equation for x = 1.
  | 1 + 6 |
  = | 7 |
  = 7
  Right Side of Equation for x = 1.
  7
• x = -13
  Left Side of Equation for x = 1.
  | -13 + 6 |
  = | -7 |
  = 7
  Right Side of Equation for x = 1.
  7

The solutions to the given equation are x = 1 and x = -13

Matched Exercise 1: Solve the equation

Answers

Example 2: Solve the equation

Solution to Example 2:

• Given
  -2 | x/2 + 3 | - 4 = -10
  
  
• We first write the equation in the form | A | = B. Add 4 to both sides and group like terms
  -2| x/2 + 3 | = -6
  
  
• Divide both sides by -2
  | x/2 + 3 | = 3
  
  
• We now proceed as in example 1 above, the equation
  | x/2 + 3 | = 3 gives two equations.
  a) x/2 + 3 = 3
  or
  b) x/2 + 3 = -3
  
  
• Solve equation a)
  x/2 + 3 = 3
  
  
• to obtain
  x = 0
  
  
• Solve equation b)
  x/2 + 3 = -3
  
  
• to obtain
  x = -12
  
  

Check solutions:

• x = 0
  Left Side of Equation for x = 0.
  -2 | x/2 + 3 | - 4
  = -2| 3 | - 4
  = -10
  Right Side of Equation for x = 1.
  -10
• x = -12
  Left Side of Equation for x = -12.
  -2 | x/2 + 3 | - 4
  = -2 | -12/2 + 3 | - 4
  = -2 | -6 + 3 | - 4
  = -2(3) - 4
  = -10
  Right Side of Equation for x = -12.
  -10

The solutions to the given equation are x = 0 and x = -12

Matched Exercise 2: Solve the equation

Answers

Example 3: Solve the equation

Solution to Example 3:

• If 2x - 2 >= 0 which is equivalent to x >= 1, then | 2x - 2 | = 2x - 2 and the given equation becomes
  2x - 2 = x + 1
  
  
• Add 2 - x to both sides
  x = 3
  
  
• Since x = 3 satisfies the condition x >= 1, it is a solution.
• If 2x - 2 < 0 which is equivalent to x < 1, then | 2x - 2 | = -(2x - 2) and the given equation becomes
  -(2x - 2) = x + 1
  
  
• Solve for x to obtain
  x = 1/3
  
  
• Since x = 1 / 3 satisfies the condition x < 1, it is a solution.

Check solutions

• x = 3
  Left Side of Equation for x = 3.
  | 2x - 2 |
  = | 2*3 - 2 |
  = 4
  Right Side of Equation for x = 3.
  x + 1
  = 3 + 1
  = 4
• x = 1/3
  Left Side of Equation for x = 1/3.
  | 2x - 2 |
  = | 2*(1/3) - 2 |
  = 4/3
  Right Side of Equation for x = 1/3.
  x + 1
  = 4/3

The solutions to the given equation are x = 3 and x = 1/3

Matched Exercise 3:Solve the equation

Answers

Example 4: Solve the equation

Solution to Example 3:

• If x² - 4 >= 0 ,or x² >= 4, then | x² - 4 | = x² - 4 and the given equation becomes
  x² - 4 = x + 2
  
  
• Add - (x + 2) to both sides
  x² - 4 -( x + 2) = 0
  
  
• Factor the left term
  (x - 2)(x + 2) -( x + 2) = 0
  
  (x + 2)(x - 2 -1) = 0
  
  (x + 2)(x - 3) = 0
  
  
• Using the factor theorem, we can write two simpler equations
  x + 2 = 0
  or
  x - 3 = 0
  
  
• Solve the above equations for x to find two values of x that make the left side of the equation equal to zero.
  x = -2 and x = 3.
  
  
• Both values satisfy the condition x² >= 4 and are solutions to the given equation.
  x = -2 and x = 3.
  
  
• If x² - 4 < 0 ,or x² < 4, then | x² - 4 | = -(x² - 4) and the given equation becomes.
  -(x² - 4) = x + 2
  
  -(x² - 4) - ( x + 2) = 0
  
  
• Factor the left term.
  -(x - 2)(x + 2) - ( x + 2) = 0
  
  (x - 2)(x + 2) + ( x + 2) = 0
  
  (x - 2)(x + 2) + ( x + 2) = 0
  
  (x + 2)(x - 2 + 1) = 0
  
  (x + 2)(x - 1) = 0
  
  
• Two values make the left side of the above equation equal to zero
  x = -2 and x = 1.
  
  
• Only x = 1 satisfies the condition x² < 4
  

Check solutions:

• x = -2
  Right Side of Equation = | x² - 4 |
  = | (-2)² - 4 | = 0
  Left Side of Equation = x + 2 = -2 + 2 = 0
  
  
• x = 3 Left Side of Equation = | x² - 4 |
  = | 3² - 4 |
  = | 5 |
  = 5 Right Side of Equation = x + 2 = 3 + 2 = 5
• x = 1
  Left Side of Equation = | x² - 4 |
  = | 1² - 4 | = | - 3 | = 3 Right Side of Equation = x + 2 = 1 + 2 = 3

Conclusion

The solutions to the given equation are x = -2, x = 1 and x = 3.

Matched Exercise 4: Solve the equation

Answers

Exercises.(see answers below)

Solve the following absolute value equations

a) | x - 4 | = 9

b) | x ² + 4 | = 5

c) | x ² - 9 | = x + 3

d) | x + 1 | = x - 3

e) | -x | = 2

Answers to Above Exercises.

a) -5 , 13

b) -1 , 1

c) -3 , 2 , 4

d) no real solutions

e) -2 , 2

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