Transcript 1.4 Solving Absolute Value Equations

1.4 Solving Absolute Value
Equations
Evaluate and solve Absolute Value
problems
Absolute Value
The absolute value of a number is its
distance from Zero on the number
line.
Distance is never negative, so absolute
value of a number is never negative
Lets look at a number line
If |x| = 4, then x is 4 units from zero,
to the left and right side. Thus x = 4
and x = -4
-4
0
4
Lets Evaluate an expression
2.7 + |6 – 2x| if x = 4
Lets Evaluate an expression
2.7 + |6 – 2x| if x = 4
2.7 + |6 – 2(4)|
Lets Evaluate an expression
2.7 + |6 – 2x| if x = 4
2.7 + |6 – 2(4)|
2.7 + |6 – 8|
Lets Evaluate an expression
2.7 + |6 – 2x| if x = 4
2.7 + |6 – 2(4)|
2.7 + |6 – 8|
2.7 + | - 2|
Lets Evaluate an expression
2.7 + |6 – 2x| if x = 4
2.7 + |6 – 2(4)|
2.7 + |6 – 8|
2.7 + | - 2| = 2.7 + 2 = 4.7
Lets solve an Equation
|y+3|=8
So
y + 3 = 8 and y + 3 = - 8
Why two equations?
Lets solve an Equation
|y+3|=8
So
y + 3 = 8 and y + 3 = - 8
y=5
and
y = - 11
|6 – 4a| + 5 = 0
Get the Absolute Value alone
So
|6 – 4a| = -5
Is there a problem with this Problem?
|6 – 4a| + 5 = 0
Get the Absolute Value alone
So
|6 – 4a| = -5
Is there a problem with this Problem?
Yes, Absolute Value can not be negative
Then No Solution
|8 + y| = 2y - 3
Here we have to break up the equation
to remove the absolute value
8 + y = + (2y – 3)
and 8 + y = - (2y – 3)
We need to distribute and negative one
in the second equation
8 + y = + (2y – 3) and 8 + y = - (2y – 3)
Given us
8 + y = 2y – 3
and
8 + y = -2y + 3
8 + y = + (2y – 3) and 8 + y = - (2y – 3)
Given us
8 + y = 2y – 3
11 = y
and
8 + y = -2y + 3
3y = - 5
y = - 5/3
Do the answers work?
|8 + y| = 2y – 3
Try y =11
Yes it works
|8 + 11| = 2(11) – 3
|19| = 22 – 3
19 = 19
Do the answers work?
|8 + y| = 2y – 3
Try y =-5/3
|8 + (-5/3)| = 2(-5/3) – 3
|19/3| = -10/3 – 3
19/3 = -19/3 Which is wrong
-5/3 does not work
So it is just y = 11
Homework
Page 30-31 #21, 23, 29 – 43 odd, 49