MATH 31 U1A LESSON #4 SOLVING ABSOLUTE VALUE

MATH 31
U1A LESSON #4
SOLVING ABSOLUTE VALUE EQUATIONS
NAME ANSWERS
(πŸ•, πŸ—)
(βˆ’πŸ, πŸ—)
Practice
1. |2π‘₯ βˆ’ 5| = 9
πŸπ’™ βˆ’ πŸ“ = πŸ—
x=7
βˆ’(πŸπ’™ βˆ’ πŸ“) = πŸ—
βˆ’πŸπ’™ + πŸ“ = πŸ—
βˆ’πŸπ’™ = πŸ’
x= –2
2. |1 βˆ’ 3π‘₯| = 7
𝟏 βˆ’ πŸ‘π’™ = πŸ•
x= –2
βˆ’(𝟏 βˆ’ πŸ‘π’™) = πŸ•
βˆ’πŸ + πŸ‘π’™ = πŸ•
πŸ‘π’™ = πŸ–
πŸ–
x= πŸ‘
(βˆ’πŸ, πŸ•)
πŸ–
( , πŸ•)
πŸ‘
3. |4π‘₯ βˆ’ 1| + 5 = 8
|πŸ’π’™ βˆ’ 𝟏| = πŸ‘
πŸ’π’™ βˆ’ 𝟏 = πŸ‘
πŸ’π’™ = πŸ’
x= 1
βˆ’(πŸ’π’™ βˆ’ 𝟏) = πŸ‘
βˆ’πŸ’π’™ + 𝟏 = πŸ‘
βˆ’πŸ’π’™ = 𝟐
𝟏
x = βˆ’πŸ
U1A L4 ANS Abs Value Equations
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𝟏
(βˆ’ , πŸ–)
𝟐
(𝟏, πŸ–)
MATH 31
U1A LESSON #4
SOLVING ABSOLUTE VALUE EQUATIONS
NAME ANSWERS
4. 1 βˆ’ |π‘₯ + 3| = βˆ’2
πŸ‘=𝒙+πŸ‘
𝒙=𝟎
(𝟎, βˆ’πŸ)
(βˆ’πŸ”, βˆ’πŸ)
πŸ‘ = |𝒙 + πŸ‘|
πŸ‘ = βˆ’(𝒙 + πŸ‘)
πŸ‘ = βˆ’π’™ βˆ’ πŸ‘
𝒙 = βˆ’πŸ”
5. |2π‘₯ βˆ’ 4| = 0
πŸπ’™ βˆ’ πŸ’ = 𝟎
πŸπ’™ = πŸ’
x= 2
(𝟐, 𝟎)
6. |3π‘₯ βˆ’ 2| + 5 = 3
|πŸ‘π’™ βˆ’ 𝟐| = βˆ’πŸ
No Solution
U1A L4 ANS Abs Value Equations
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MATH 31
U1A LESSON #4
SOLVING ABSOLUTE VALUE EQUATIONS
NAME ANSWERS
Practice
7. |3π‘₯ + 5| = 2π‘₯
πŸ‘π’™ + πŸ“ = πŸπ’™
𝒙 = βˆ’πŸ“
βˆ’(πŸ‘π’™ + πŸ“) = πŸπ’™
𝒙 = βˆ’πŸ
CHECK
|3(βˆ’πŸ“) + 5| = 2(βˆ’πŸ“)
10 β‰  βˆ’10
CHECK
|3(βˆ’πŸ) + 5| = 2(βˆ’πŸ)
2 β‰  βˆ’2
There is NO solution
8. |1 βˆ’ π‘₯| = 3π‘₯ βˆ’ 2
𝟏 βˆ’ 𝒙 = πŸ‘π’™ βˆ’ 𝟐
βˆ’πŸ’π’™ = βˆ’πŸ‘
πŸ‘
𝒙=
πŸ’
CHECK
πŸ‘
πŸ‘
|1 βˆ’ | = 3 ( ) βˆ’ 2
πŸ’
πŸ’
1 1
=
4 4
𝟏 βˆ’ 𝒙 = βˆ’(πŸ‘π’™ βˆ’ 𝟐) = βˆ’πŸ‘π’™ + 𝟐
πŸπ’™ = 𝟏
𝟏
𝒙=
𝟐
CHECK
𝟏
𝟏
|1 βˆ’ | = 3 ( ) βˆ’ 2
𝟐
𝟐
1
1
β‰ βˆ’
2
2
πŸ‘ 1
( , )
πŸ’ 4
1
9. |π‘₯ βˆ’ 2| = βˆ’ 2 π‘₯ + 3
𝟏
π’™βˆ’πŸ=βˆ’ 𝒙+πŸ‘
𝟐
πŸ‘
𝒙=πŸ“
𝟐
𝟐 𝟏𝟎
𝒙 = πŸ“× =
πŸ‘
πŸ‘
CHECK
𝟏𝟎
1 𝟏𝟎
| βˆ’ 2| = βˆ’ ( ) + 3
πŸ‘
2 πŸ‘
4
10 18
=βˆ’ +
3
6
6
4 4
=
3 3
U1A L4 ANS Abs Value Equations
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𝟏
𝟏
𝒙 βˆ’ 𝟐 = βˆ’ (βˆ’ 𝒙 + πŸ‘) = 𝒙 βˆ’ πŸ‘
𝟐
𝟐
𝟏
𝒙 = βˆ’πŸ
𝟐
𝟐
𝒙 = βˆ’πŸ× = βˆ’πŸ
𝟏
CHECK
(βˆ’πŸ, 4)
1
|βˆ’πŸ βˆ’ 2| = βˆ’ (βˆ’πŸ) + 3
2
4=4
𝟏𝟎 4
( , )
πŸ‘ 3
MATH 31
U1A LESSON #4
SOLVING ABSOLUTE VALUE EQUATIONS
NAME ANSWERS
10. |π‘₯ + 6| = |2π‘₯|
(πŸ”, 12)
𝒙 + πŸ” = πŸπ’™
πŸ”=𝒙
𝒙 + πŸ” = βˆ’πŸπ’™
πŸ‘π’™ = βˆ’πŸ”
𝒙 = βˆ’πŸ
(βˆ’πŸ, 4)
11. |1 βˆ’ 3π‘₯| = |π‘₯ βˆ’ 7|
𝟏 βˆ’ πŸ‘π’™ = 𝒙 βˆ’ πŸ•
βˆ’πŸ’π’™ = βˆ’πŸ–
𝒙=𝟐
𝟏 βˆ’ πŸ‘π’™ = βˆ’π’™ + πŸ•
βˆ’πŸπ’™ = πŸ”
𝒙 = βˆ’πŸ‘
(βˆ’πŸ‘, 10)
(𝟐, 5)
CHECK
|1 βˆ’ 3(𝟐)| = |𝟐 βˆ’ 7|
CHECK
|1 βˆ’ 3(βˆ’πŸ‘)| = |βˆ’πŸ‘ βˆ’ 7|
5=5
10 = 10
U1A L4 ANS Abs Value Equations
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