Solving Absolute Value Inequalities, Slides of Algebra

Download Solving Absolute Value Inequalities and more Slides Algebra in PDF only on Docsity!

§4.3b AbsVal

InEqualities

Review §

 Any QUESTIONS About

• §4.3a → Absolute Value

 Any QUESTIONS About HomeWork

• §4.3a → HW-

4.3 MTH 55

Example  AbsVal & <

• Given InEquality: | x − 3| < 6
  • solve, graph the solution set, and write the solution set in both set-builder and interval notation
• SOLUTION
  • | x – 3| < 6 → x − 3 > −6 and x − 3 < 6
  • thus −6 < x − 3 < 6
    • ReWritten as Compound InEquality
  • So −3 < x < 9 (add +3 to all sides)

Example  AbsVal & <

• SOLUTION: | x − 3| < 6
  • Thus the Solution → −3 < x < 9
  • Solution in Graphical Form

-10-10 -9 -8 -7 -6 -5 -4 -3^ (^ -2 -1 0 1 2 3 4 5 6 7 8 ) 9 1010

• Set-builder notation: { x | −3 < x < 9}
• Interval notation: (−3, 9)

Example  |2 x − 3| + 8 < 5

• The InEquality Simplified to:

|2 x − 3| < − 3

• Since the absolute value cannot be less than a negative number, this inequality has NO solution: Ø - No Graph - Set-Builder Notation → {Ø} - Interval notation: We do not write interval notation because there are no values in the solution set.

Solving AbsVal InEqual with >

• Solving Inequalities in the Form | x | > a , where a > 0

1. Rewrite as a compound inequality involving “or”: x < −a OR x > a.
2. Solve the compound inequality

• Similarly, to solve | x | ≥ a , we would write x ≤ −a or x ≥ a

Example  AbsVal & >

• SOLUTION : | x + 7| > 5
  • x + 7 < − 5 or x + 7 > 5
    • The Addition Principle Produces Solutions
  • x < − 12 or x > − 2
  • The Graph

-15-15 -14 -13 -12^ )^ -11 -10 -9 -8 -7 -6 -5 -4 -3 -2 ( -1 0 1 2 3 4 55

• Set-builder notation: { x| x < −2 or x > −2}
• Interval notation: (−∞, −12) U (−2, ∞).

Example  |4 x + 7| – 9 > –

• Given InEquality: |4 x + 7| − 9 > − 12
  • solve, graph the solution set, and write the solution set in both set-builder and interval notation
• SOLUTION
• Isolate the absolute value
  • |4 x + 7| – 9 > –
  • |4 x + 7| > –3 ???

Summary: Solve | ax + b | > k

• Let k be a positive real number, and p and q be real numbers.
• To solve | a x + b | > k , solve the following compound inequality a x + b > k OR a x + b < − k.
• The solution set is of the form (−∞, p )U( q , ∞), which consists of two Separate intervals. p q

Example  |2 x + 3| > 5

• By the Previous Slide this absolute value inequality is rewritten as 2 x + 3 > 5 or 2 x + 3 < −
• The expression 2 x + 3 must represent a number that is more than 5 units from 0 on either side of the number line. - Use this analysis to solve the compound inequality Above

Summary: Solve | ax + b | < k

• Let k be a positive real number, and p and q be real numbers.
• To solve | a x + b | < k , solve the three-part “and” inequality - k < ax + b < k
• The solution set is of the form ( p , q ), a single interval.

p q

Example  |2 x + 3| < 5

• By the Previous Slide this absolute value inequality is rewritten as − 5 < 2 x + 3 and 2 x + 3 < 5
• In 3-Part form

− 5 < 2 x + 3 < 5

• Solving for x

−5 < 2 x + 3 < 5 −8 < 2 x < 2 −4 < x < 1

Caution for AbsVal vs <>

• When solving absolute value inequalities of the types > & < remember the following:

1. The methods described apply when the constant is alone on one side of the equation or inequality and is positive.
2. Absolute value equations and absolute value inequalities of the form | ax + b | > k translate into “ or ” compound statements.

Caution for AbsVal vs <>

• When solving absolute value inequalities of the types > & < remember the following:

3. Absolute value inequalities of the form | ax + b | < k translate into “ and ” compound statements, which may be written as three-part inequalities.
4. An “or” statement cannot be written in three parts. It would be incorrect to use −5 > 2 x + 3 > 5 in the > Example, because this would imply that − 5 > 5, which is false