Unit 2 Section 10 Absolute Value Equations.notebook November 05, 2015 Nov 5­9:54 AM

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Unit 2 Section 10 Absolute Value Equations.notebook
Nov 5­9:54 AM
Nov 5­9:54 AM
November 05, 2015
Unit 2 Section 10 Absolute Value Equations.notebook
Nov 5­9:54 AM
Absolute Value Equations
Unit 2
Oct 13­8:49 AM
November 05, 2015
Unit 2 Section 10 Absolute Value Equations.notebook
November 05, 2015
Absolute Value Equations
Example 1:
|x| = 2 x = 2 or x = ­2
x is a number whose distance is 2 units away from the origin.
Example 2:
|x| + 3 = 8
Oct 13­8:50 AM
The equation |z ­ 5| = 6 means that the distance on a number line from z ­ 5 to 0 is 6 units. There are two points that are 6 units from 0: 6 and ­6. So to find the values of x, solve the equations z ­ 5 = 6 and z ­ 5 = ­6. Example 3: |z ­ 5| = 6
Oct 13­8:52 AM
Unit 2 Section 10 Absolute Value Equations.notebook
November 05, 2015
The absolute value of an expression cannot be negative, so there is no solution.
1.) | t + 6| = ­4
2.) |2y + 1| = ­6
Oct 13­8:53 AM
Always isolate the absolute value before splitting into two equations.
Try these:
3.) 3 = |w| ­ 4
4.) |x| + 5 = 4 ­ 6
Oct 13­8:53 AM
Unit 2 Section 10 Absolute Value Equations.notebook
November 05, 2015
6.) 3|3x ­ 7| + 6 = 9
5.) 4|n| = 32
Oct 13­8:52 AM
7.) 7|3x ­ 5| = ­14
Oct 13­8:56 AM
Unit 2 Section 10 Absolute Value Equations.notebook
8.) |x + 6 | = 0
November 05, 2015
9.) |2x ­ 1| = 3x
Nov 8­11:00 AM
10.) |3y ­ 2| = 2y ­ 1
Nov 8­11:02 AM
Unit 2 Section 10 Absolute Value Equations.notebook
November 05, 2015
Bell Ringer
1.) |2z ­ 1| + 2 = 3z + 4
Nov 4­8:57 AM
2.) Bonnie estimates her stride to be 16 in. However, any given stride is likely to vary from her estimate up to 2 in. Write and solve an equation to find Bonnie's maximum and minimum stride.
Oct 13­8:57 AM
Unit 2 Section 10 Absolute Value Equations.notebook
November 05, 2015
3.) The ideal circumference of a woman's basketball is 28.75 in. The actual circumference may vary from the ideal by at most 0.25 in. What are the acceptable circumferences for a women's basketball?
Nov 5­10:57 AM
4.) The absolute value of twice a number added to 5 is 33.
5.) Five times the absolute value of the difference of four times a number and 4 is 20.
6.) If 4 is subtracted from twice the absolute value of the difference of twice a number and 2, the result is 24.
Nov 8­11:03 AM
Unit 2 Section 10 Absolute Value Equations.notebook
Homework:
Homework Packet
page 29 and 30
Nov 4­9:05 AM
Oct 9­8:46 AM
November 05, 2015
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