1. 1 Standards 2, 5, 25 SOLVING TWO VARIABLE LINEAR INEQUALITIES INCLUDING ABSOLUTE VALUE INEQUALITIES PROBLEM 1 PROBLEM 3 PROBLEM 4 PROBLEM 2 PROBLEM 5 PROBLEM 6 PROBLEM 7 PROBLEM 9 PROBLEM 10 PROBLEM 8 END SHOW SO, WHEN WE SHADE ABOVE OR BELOW? MOVING AROUND AN ABSOLUTE VALUE GRAPH PROBLEM 11 PROBLEM 12 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2. 2 Standard 2: Students solve systems of linear equations and inequalities (in two or three variables) by substitution, with graphs, or with matrices. Estándar 2: Los estudiantes resuelven sistemas de ecuaciones lineares y desigualdades (en 2 o tres variables) por substitución, con gráficas o con matrices. Standard 5: Students demonstrate knowledge of how real and complex numbers are related both arithmetically and graphically. In particular, they can plot complex numbers as points in the plane. Estándar 5: Los estudiantes demuestran conocimiento de como los números reales y complejos se relacionan tanto aritméticamente como gráficamente. En particular, ellos grafican números como puntos en el plano. Standard 25: Students use properties from number systems to justify steps in combining and simplifying functions. Estándar 25: Los estudiantes usan propiedades de sistemas numéricos para justificar pasos en combinar y simplificar funciones. ALGEBRA II STANDARDS THIS LESSON AIMS: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3. 3 Standard 5 Graph the following inequality: x > 2 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y We shaded at the right of the line because x is more than 2. The line is dashed because it is not equal or less than x, so the line which is the boundary is not included in the solution. x = 2 Boundary is: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4. 4 Standard 5 Graph the following inequality: m= 0 y- intercept = (0,-2) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y We shaded above the line because y is more or equal than -2. The line is continuous because the word equal, so the line which is the boundary is included in the solution. y = 0x – 2 Boundary is: y - 2 > PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5. 5 Standard 5 Graph the following inequality: y < 6 m= 0 y- intercept = (0,6) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y We shaded below the line because y is less than 6. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included in the solution. y = 0x + 6 Boundary is: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6. 6 Standard 5 Graph the following inequality: 4x + 2y < 10 -4x 2y < -4x + 10 2 y < - x + 4 2 10 2 y < -2x + 5 m= -2 y- intercept = (0,5) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y +1 2 1 + - = - 2 We shaded below the line because y is less than the expression -2x + 5. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included. y = -2x + 5 Boundary is: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7. 7 Standard 5 Graph the following inequalitie: -9x + 3y< 3 +9x 3y< 9x + 3 3 y < x + 9 3 3 3 y < 3x + 1 m= 3 y- intercept = (0, 1) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y 3 1 + + = 3 + +1 We shaded below the line because y is less than the expression 3x +1. The line is dashed because it is not equal or less than y, so the line which is the boundary is not included. Boundary is: y = 3x + 1 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

8. 8 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y +2 y- intercept = (0,-3) m 5 4 + + = +4 5 + y – 2 (x – 4) 5 4 > y - 2 x - 20 4 5 4 > y - 2 x - 5 5 4 > y x -3 5 4 > Graph the following inequality: Boundary is: y = x -3 5 4 We shaded above because y is greater or equal than the expression and the line is continuous because the word equal in greater or equal indicates that the boundary is included in the solution. x -3 5 4 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9. 9 Standard 5 Summarizing: We shade above the line We shade below the line Continuous line Dashed line PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10. 10 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following absolute value equation: y = |x| For x < 0 y = -x y = x For x 0 > Now let’s shift it two units above: y = |x| + 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11. 11 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following absolute value equation: y = |x| For x < 0 y = -x y = x For x 0 > Now let’s shift it two units above: y = |x| + 2 Now let’s shift it three units to the right: y = |x-3| + 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

12. 12 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following absolute value equation: y = |x| For x < 0 y = -x y = x For x 0 > Now let’s shift it two units above: y = |x| + 2 Now let’s shift it three units to the right: y = |x-3| + 2 Now let’s graph it upside down y = – |x-3| + 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13. 13 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following absolute value equation: y = |x| For x < 0 y = -x y = x For x 0 > Now let’s shift it two units above: y = |x| + 2 Now let’s shift it three units to the right: y = |x-3| + 2 Now let’s graph it upside down y = – |x-3| + 2 Now let’s make it skinner y = – 6|x-3| + 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14. 14 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following absolute value equation: y = |x| For x < 0 y = -x y = x For x 0 > Now let’s shift it two units above: y = |x| + 2 Now let’s shift it three units to the right: y = |x-3| + 2 Now let’s graph it upside down y = – |x-3| + 2 Now let’s make it skinner y = – 2|x-3| + 2 So, we have learn how the different parameters in an absolute value equation affect our graph. Now let’s graph absolute value inequalities. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15. 15 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following inequality: y > |x| For x < 0 y = -x y = x For x 0 > Finding the boundary: Testing (0,2) There are two regions: 2 > | 0| 2 > 0 true Therefore, the region where (0,2) lies is the solution region and we shade it. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16. 16 Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following inequality: y < |x+1|– 3 For x < 0 y = -(x+1) – 3 y = x+1 – 3 For x 0 > Finding the boundary: Testing (0,0) There are two regions: 0 < | 0+1| – 3 0 < -2 false So the region where (0,0) lies is not in the solution region, therefore we shade the region below. y = - x – 1 -3 y = - x – 4 y = x – 2 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17. Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following inequality: |y| > 2x+1 Finding the boundary: |y| = 2x+1 For this equation to yield real number values it has to be greater or equal to zero: 2x+1 0 > 2 2x -1 > And, we explore around this value: |y| = 2( )+1 = 0 1 2 1 2 - 0 1 3 5 |y| = -1 |y| = 0 |y| = 1 |y| = 3 |y| = 5 0 -1, 1 -3, 3 -5, 5 0 1 2 1 2 - (, 0) 1 2 - (0,-1), (0,1) (1,-3), (1,3) (2,-5), (2,5) x |y| = 2x+1 y (x,y) Testing (1,0) There are two regions: | | > 2( )+1 0 1 0 > 3 x > 1 2 - false So we shade the other region. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18. Standard 5 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y Graph the following inequality: |y| < |x–3 | Finding the boundary: |y| = |x–3 | x–3 = 0 +3 x = 3 Then, we explore around this value: |y| = |( )–3| = 2 1 |y| = 1 -1, 1 2 (2,-1), (2,1) x |y| = |x–3| y (x,y) Testing (0,0) | | < |( )–3| 0 0 0 < 3 true Therefore, these are the solution regions. |y| = |( )–3| = 3 0 |y| = 0 0 3 (3,0) |y| = |( )–3| = 4 1 |y| = 1 -1, 1 4 (4,-1), (4,1) Testing (4,0) | | < |( )–3| 0 4 0 < 1 true |x–3| is always greater or equal than zero, so: PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19. Standard 5 Graph the following inequality: Finding the boundary: For this equation to yield real number values it has to be greater or equal to zero: Thus, we explore around this domain: |y| = -|( )+2|+2 = -5 |y| = -1 - 5 |y| = -|( )+2|+2 = -4 0 |y| = 0 0 - 4 (-4,0) |y| = -|( )+2|+2 = -3 1 |y| = 1 -1, 1 - 3 (-3,-1), (-3,1) |x+2| + |y| 2 > |x+2| + |y| = 2 -|x+2| |y| = -|x+2| +2 -|x+2| + 2 0 > -2 -|x+2| -2 > |x+2| 2 < x +2 2 < x +2 -2 > x | -4 x 0 < < x y (x,y) |y| = -|x+2| +2 |y| = -|( )+2|+2 = -2 2 |y| = 2 -2, 2 - 2 (-2,-2), (-2,2) |y| = -|( )+2|+2 = 1 |y| = 1 -1, 1 - 1 (-1,-1), (-1,1) |y| = -|( )+2|+2 = 1 |y| = -1 1 |y| = -|( )+2|+2 = 0 0 |y| = 0 0 0 (0,0) 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y There are two regions, we test one of them: Using (-2,0) |-2+2| + |0| 2 > 0 2 > False, so the outer area is the solution. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

20. Standard 5 Graph the following inequality: 4 2 6 -2-4-6 2 4 6 -2 -4 -6 8 10 -8 -10 8 -8 10 x y |2x+y| 2 < or -2x y -2x + 2 < > y = -2x + 2 y = -2x – 2 These are the boundary lines: Testing (0,0) |2(0)+(0)| 2 < 0 2 < True, so the area in between the lines is the solution, including the lines. 2x + y 2< 2x + y -2 > PRESENTATION CREATED BY SIMON PEREZ. All rights reserved