Assignment 1.4 Solve Absolute Value Equations  x + 11 = 42

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Assignment 1.4
Solve Absolute Value Equations
Solve each equation. Check your solution.
 x + 11 = 42
Since 42 ≥ 0, solutions exists
x + 11 = 42
-11 -11
or
x + 11 = -42
-11 = -11
42 – 11 =
x
or
x
-42-11 =
=
=
3  x + 6  = 36
3
3
x+6=
36 ÷ 3 =
13
Since 12 ≥ 0, solutions exist
x+6=
-6
13
-6
x =
or x + 6 = -12
-6 = -6
x
13
-6=
=
-12 – 6 =
 4x - 5 = -25
since -25 ___ 0, there are No Solutions.
a) <
b) >
Closed Last Step Strategy
13
Assignment 1.4
Solve Absolute Value Equations
 x + 7  = 3x – 5
since it is not known if 3x – 5 ≥ 0, the answers need to be checked.
x + 7 = 3x – 5
-x
-x
7 = 2x -5
+5
+5
12
2x
2
2
or
x+7
x+7
+3x
= - (3x – 5)
= - 3x + 5
+3x
4x+ 7 =
-7
4x
4
5
-7
=
-2
4
12 ÷ 2 =
=x
x =
(3(6)-5)≥0
so, x=
since 3(- ½ )-5<0
is a solution
x = - ½ is not a solution
-2 ÷ 4 =
=x
 y – 5  - 2 = 10
y–5
+ 2 = +2
12
Since 12 ≥ 0, there are solutions
y – 5 = 12
+ 5 +5
or
y
or
=
y – 5 = -12
+5 +5
y =
Closed Last Step Strategy
12 + 5 =
-12 +5 =
Assignment 1.4
Solve Absolute Value Equations
4  3x + 4  = 4x + 8
4
4 4
 3x + 4  = x + 2
Since it is not known if x +2 ≥ 0, the answers must be checked
3x + 4 = x + 2
-x
-x
2x + 4
2
-4
-4
2x
2
or
3x + 4 = - (x +2)
3x + 4 = -x – 2
+x
+x
4x + 4
-2
-4
-4
-2
2
4x
4
-6
4
x=
-2 ÷ 2 =
-6 - 3
x= 4
2
since -1 + 2 ≥ 0,
x = -1 is a solution
since -3/2 + 2 ≥ 0, x = -3/2 is a solution
or x = -3
2
x=
= -1 ½
BIKING
Paloma’s training goal is to ride four miles on her bicycle in 15 minutes. If her actual time is always
within plus or minus 3 minutes of her preferred time, how long are her shortest and longest rides?
 x – 15  = 3
x - 15
=
+ 15
x
3
or
+15
=
x - 15
=-3
+ 15
or
x
+15
=
Closed Last Step Strategy
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