Solve equations with absolute value; including examples and questions with detailed solutions and explanations.

Review of Absolute Value

Examples with Solutions

• If |x + 6 | = 7, then (see rule 5 above)
  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
  

• 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

Solution to Matched Exercise

• 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
  

• 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

Solution to Matched Exercise

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

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

Solution to Matched Exercise

• 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
  

• 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

Solution to Matched Exercise

Solutions to the Above Matched Exercises

Matched Exercise 1

Matched Exercise 2

Matched Exercise 3

Matched Exercise 4

More Exercises with Answers

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