1.4 Solving Absolute Value Equations Evaluate and solve Absolute Value problems.

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Presentation transcript:

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 =

Lets Evaluate an expression |6 – 2x| if x = 4

Lets Evaluate an expression |6 – 2x| if x = |6 – 2(4)|

Lets Evaluate an expression |6 – 2x| if x = |6 – 2(4)| |6 – 8|

Lets Evaluate an expression |6 – 2x| if x = |6 – 2(4)| |6 – 8| | - 2|

Lets Evaluate an expression |6 – 2x| if x = |6 – 2(4)| |6 – 8| | - 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 = 5andy = - 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 and 8 + y = -2y = y 3y = - 5 y = - 5/3

Do the answers work? |8 + y| = 2y – 3 Try y =11|8 + 11| = 2(11) – 3 |19| = 22 – 3 19 = 19 Yes it works

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 #21, 23, 29 – 43 odd, 49