Tues, 3/5 SWBAT… solve absolute value equations Agenda WU (5 min) Binder check (10 min) 5 Examples: absolute value equations (25 min) Warm-Up: Place your test corrections and get ready for the next unit on your desk. Take out your binder. Set-up notes. Topic = “Solving Absolute Value Equations.” HW#1: Absolute value equations 1 1

New unit on Absolute Value & Inequalities (3 weeks) Daily HW, WU, exit slips, weekly quizzes SWBAT… Solve absolute value equations. Solve and graph one-step inequalities involving addition, subtraction, multiplication, and division. Solve and graph multi-step inequalities. Write inequalities in interval notation. Solve and graph compound inequalities. Graph inequalities with two variables. Graph system of inequalities. Unit test on Friday, March 22 (before Spring Break)

Solving Absolute Value Equations

1 block 1 block You walk directly east from your house one block. How far from your house are you? You walk directly west from your house one block. How far from your house are you? It didn't matter which direction you walked, you were still 1 block from your house. This is like absolute value. It is the distance from zero. It doesn't matter whether we are in the positive direction or the negative direction, we just care about how far away we are. 4 units away from 0 4 units away from 0 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8

Using Absolute Value in Real Life The graph shows the position of a diver relative to sea level. Use absolute value to find the diver’s distance from the surface.

Definition of Absolute Value The distance from any number to zero on the number line. The value is always positive. Why? Because absolute value is a distance and distance is always positive. Distance is always positive because absolute value is a distance and distance is always positive. Odometer on a car always begins at zero… it is never negative. No such thing as a negative one mile run. Even if you walk backwards 3ft, and measure from beginning to ending point, the yard stick will still measure 3ft, not -3ft.

To solve an absolute value equation: Ex. #1 To solve an absolute value equation: Isolate the absolute value on one side of the equal sign. Case 1: Set the inside of the absolute value equal to a positive of the other given expression. Solve. Case 2 : Set the inside the absolute value equal to the negative of the other given expression. Solve. Check both solutions. |x| = 6 x = 6 or x = -6 Check: |x | = 6 or |x| = 6 | 6 | = 6 | -6 | = 6 6 = 6 6 = 6

6 and -6 are both 6 units away from 0 What we are after here are values of x such that they are 6 away from 0. 2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 4 6 8 6 and -6 are both 6 units away from 0

To solve an absolute value equation: Ex. #2 To solve an absolute value equation: Isolate the absolute value on one side of the equal sign. Case 1: Set the inside of the absolute value equal to a positive of the other given expression. Solve. Case 2 : Set the inside the absolute value equal to the negative of the other given expression. Solve. Check both solutions. |x + 3| = 7 x + 3 = 7 or x + 3 = -7 Subtract 3 from both sides x = 4 or x = -10 Check: |x + 3| = 7 or |x + 3| = 7 | 4 + 3| = 7 | -10 + 3| = 7 |7| = 7 |-7| = 7 7 = 7 7 = 7

Ex. #3 |15 – 3x| = 6 15 – 3x = 6 or 15 – 3x = -6 . -3x = -9 -3x = -21 Subtract 15 from both sides. x = 3 or x = 7 Divide both sides by –3. Check: |15 – 3x| = 6 |15 – 3x| = 6 |15 – 3(3)| = 6 |15 – 3(7)| = 6 |6| = 6 |–6| = 6 6 = 6 6 = 6

| x | = 3 Add 6 to both sides x = 3 or x = -3 Ex. #4 | x | – 6 = -3 Check: |x | - 6 = -3 or |x| - 6 = -3 | 3 | = 3 | -3 | = 3 3 = 3 3 = 3

-3c = 6 or -3c = -6 Divide both sides by -3 Ex. #5 │-3c│ – 10 = -4 │-3c│ = 6 Add 10 to both sides -3c = 6 or -3c = -6 Divide both sides by -3 c = -2 or c = 2

Ex. #6 2| x | = -10 | x | = -5 No Solution

Absolute Value and No Solutions Absolute value is always positive (or zero).  An equation such as │x │= -5 or │x – 4│= -6  is never true.  It has NO Solution.  │x │= -5 has no solution This is a distance And this is negative Ever heard of a negative distance?

OYO (On Your Own) Solve: 1.) │2x + 4│ = 12 2.) 3│x│= 6 3.) │2x + 4│ – 12 = -12

Exit Slip │2x + 4│ = -12 Solve for c and check: │3c│ – 45 = -18