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Chapter 1: Tools of Algebra 1-5: Absolute Value Equations and Inequalities Essential Question: What is the procedure used to solve an absolute value equation of inequality? (Tomorrow)
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1-5: Absolute Value Equations and Inequalities A BSOLUTE V ALUE E QUATIONS HAVE TWO SOLUTIONS, because the quantity inside the absolute value sign can be positive or negative Like compound inequalities, create two equations, and solve them independently. 1. G ET THE ABSOLUTE VALUE PORTION ALONE 2. S ET THE ABSOLUTE VALUE PORTION EQUAL TO BOTH THE POSITIVE AND NEGATIVE
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1-5: Absolute Value Equations and Inequalities Example: Solve |2y – 4| = 12 2y – 4 = 122y – 4 = -12
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1-5: Absolute Value Equations and Inequalities Example: Solve |2y – 4| = 12 2y – 4 = 122y – 4 = -12 +4 +4 2y = 16 +4 +4 2y = -8
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1-5: Absolute Value Equations and Inequalities Example: Solve |2y – 4| = 12 y = 8 or y = -4 Check: |2(8) – 4| = |16 – 4| = |12| = 12 |2(-4) – 4| = |-8 – 4| = |-12| = 12 2y – 4 = 122y – 4 = -12 +4 +4 2y = 16 2 y = 8 +4 +4 2y = -8 2 y = -4
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1-5: Absolute Value Equations and Inequalities Multiple Step Absolute Value Equations Example 2: Solve 3|4w – 1| – 5 = 10 Get the absolute value portion alone 3|4w – 1| – 5 = 10
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1-5: Absolute Value Equations and Inequalities Multiple Step Absolute Value Equations Example 2: Solve 3|4w – 1| – 5 = 10 Get the absolute value portion alone 3|4w – 1| – 5 = 10 + 5 +5 3|4w – 1| = 15
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1-5: Absolute Value Equations and Inequalities Multiple Step Absolute Value Equations Example 2: Solve 3|4w – 1| – 5 = 10 Get the absolute value portion alone 3|4w – 1| – 5 = 10 + 5 +5 3|4w – 1| = 15 3 3 |4w – 1| = 5 Now we can split into two equations, just like the last problem
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1-5: Absolute Value Equations and Inequalities |4w – 1| = 5 4w – 1 = 54w – 1 = -5
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1-5: Absolute Value Equations and Inequalities |4w – 1| = 5 4w – 1 = 54w – 1 = -5 +1 +1 4w = 6 +1 +1 4w = -4
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1-5: Absolute Value Equations and Inequalities |4w – 1| = 5 w = 1.5 or w = -1 Check (use the original problem): 3|4(1.5) – 1| – 5 = 3|6 – 1| – 5 = 3|5| – 5 = 3(5) – 5 = 15 – 5 = 10 3|4(-1) – 1| – 5 = 3|-4 – 1| – 5 = 3|-5| – 5 = 3(5) – 5 = 15 – 5 = 10 4w – 1 = 54w – 1 = -5 +1 +1 4w = 6 4 w = 1.5 +1 +1 4w = -4 4 w = -1
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1-5: Absolute Value Equations and Inequalities Checking for Extraneous Solutions Sometimes, we’ll get a solution algebraically that fails when we try and check it. These solutions are called extraneous solutions. Example 3: Solve |2x + 5| = 3x + 4 Is the absolution value portion alone? Yes When we split this into two equations, we have to NEGATE THE ENTIRE RIGHT SIDE OF THE EQUATION
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1-5: Absolute Value Equations and Inequalities |2x + 5| = 3x + 4 2x + 5 = 3x + 42x + 5 = -3x – 4
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1-5: Absolute Value Equations and Inequalities |2x + 5| = 3x + 4 2x + 5 = 3x + 42x + 5 = -3x – 4 -5 -5 2x = 3x – 1 -5 -5 2x = -3x – 9
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1-5: Absolute Value Equations and Inequalities |2x + 5| = 3x + 4 2x + 5 = 3x + 42x + 5 = -3x – 4 -5 -5 2x = 3x – 1 -3x -x = -1 -5 -5 2x = -3x – 9 +3x 5x = -9
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1-5: Absolute Value Equations and Inequalities |2x + 5| = 3x + 4 x = 1 or x = -1.8 You’ll have to check your solutions (next slide) 2x + 5 = 3x + 42x + 5 = -3x – 4 -5 -5 2x = 3x – 1 -3x -x = -1 -1 x = 1 -5 -5 2x = -3x – 9 +3x 5x = -9 5 x = -1.8
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1-5: Absolute Value Equations and Inequalities |2x + 5| = 3x + 4 x = 1 |2(1) + 5| = 3(1) + 4 |2 + 5| = 3 + 4 |7| = 7 (good) x = -1.8 |2(-1.8) + 5| = 3(-1.8) + 4 |-3.6 + 5| = -5.4 + 4 |1.4| = -1.4 (bad) The only solution is x = 1 -1.8 is an extraneous solution.
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1-5: Absolute Value Equations and Inequalities Assignment Page 36 Problems 1 – 15 (all) You will have to check your solutions for problems 10-15, so show work and identify any extraneous solutions
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Chapter 1: Tools of Algebra 1-5: Absolute Value Equations and Inequalities Day 2 Essential Question: What is the procedure used to solve an absolute value equation of inequality?
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1-5: Absolute Value Equations and Inequalities When we solved absolute value equations, we got the absolute value section alone, and set two equations One as normal One where we flipped everything outside the absolute value When solving absolute value inequalities, we do the same thing, except in addition to flipping everything on the other side of the absolute value, FLIP THE INEQUALITY AS WELL The two lines will always either split apart (greater than) or come together (less than)
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1-5: Absolute Value Equations and Inequalities Example: Solve |3x + 6| > 12. Graph the solution. 3x + 6 > 123x + 6 < -12
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1-5: Absolute Value Equations and Inequalities Example: Solve |3x + 6| > 12. Graph the solution. 3x + 6 > 123x + 6 < -12 -6 -6 3x > 6 -6 -6 3x < -18
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1-5: Absolute Value Equations and Inequalities Example: Solve |3x + 6| > 12. Graph the solution. Open circle or closed circle? Come together or split apart? 3x + 6 > 123x + 6 < -12 -6 -6 3x > 6 3 x > 2 -6 -6 3x < -18 3 x < -6
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1-5: Absolute Value Equations and Inequalities Example: Solve |3x + 6| > 12. Graph the solution. Open circle or closed circle? Closed circle (line underneath) Come together or split apart? Split apart 3x + 6 > 123x + 6 < -12 -6 -6 3x > 6 3 x > 2 -6 -6 3x < -18 3 x < -6
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1-5: Absolute Value Equations and Inequalities Solve 3|2x + 6| - 9 < 15. Graph the solution. Need to get the absolute value alone first. 3|2x + 6| - 9 < 15 +9 +9 3|2x + 6| < 24 3 3 |2x + 6| < 8
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1-5: Absolute Value Equations and Inequalities |2x + 6| < 8 2x + 6 < 82x + 6 > -8
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1-5: Absolute Value Equations and Inequalities |2x + 6| < 8 2x + 6 < 82x + 6 > -8 -6 -6 2x < 2 -6 -6 2x > -14
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1-5: Absolute Value Equations and Inequalities |2x + 6| < 8 Open circle or closed circle? Come together or split apart? 2x + 6 < 82x + 6 > -8 -6 -6 2x < 2 2 x < 1 -6 -6 2x > -14 2 x > -7
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1-5: Absolute Value Equations and Inequalities |2x + 6| < 8 Open circle or closed circle? Open circle (no line) Come together or split apart? Come together 2x + 6 < 82x + 6 > -8 -6 -6 2x < 2 2 x < 1 -6 -6 2x > -14 2 x > -7
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1-5: Absolute Value Equations and Inequalities Assignment Page 36 Problems 16 – 27 (all) Rest of week, Chapter 1 Test Wednesday: Preview Thursday: Review Friday: Test Day
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