1. Welcome to MM 212 Unit 4 Seminar!

2. Graphing and Functions

3. The Rectangular Coordinate System

4. 4 Rectangular Coordinate System A rectangular coordinate system consists of a horizontal number line and a vertical number line. x y 11 22 33 44 123 4 22 33 44 1 2 3 4 x-axis y-axis An ordered pair (x, y) represents a point on the coordinate system. (3,  4) The ordered pair (3,  4) means that x = 3 and y =  4.

5. 5 Plotting Points Example: Plot the points (4, 2), (3,  3), and (  1, 0) on a rectangular coordinate system. x y 11 22 33 44 123 4 22 33 44 1 2 3 4 (  1, 0) (4, 2) (3,  3)

6. 6 Determining Coordinates Example: Write the coordinates of each point plotted in the graph. (  4, 3) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 A B C 4 units to the left on the x-axis 3 units up a line parallel to the y-axis A = (  4, 3) B = (2, 2) C = (0,  3) (2, 2) (0,  3)

7. 7 Ordered Pairs for a Linear Equation A linear equation in two variables is an equation that can be written in the form Ax + By = C where A, B, and C are real numbers but A and B are not both zero. A solution to a linear equation is an ordered pair that makes the equation a true mathematical statement. 2x + y = 5 (2, 1) is a solution to the equation 2x + y = 5. 2(2) + (1) = 5 4 + 1 = 5 5 = 5

8. 8 Ordered Pairs for a Linear Equation Example: Find the missing coordinate to complete the following ordered-pair solution for the equation y = 6x + 5. a. (2, ?) b. (?,  7) y = 6x + 5 y = 6(2) + 5 y = 12 + 5 y = 17 y = 6x + 5  7 = 6x + 5  12 = 6x  2 = x (2, 17) is a solution. (  2,  7) is a solution.

9. 9 Solving a Formula for a Specified Value Procedure to Solve a Formula for a Specified Value 1.Remove any parentheses. 2.If fractions exist, multiply all terms on both sides by the LCD of the fractions. 3.Combine like terms, if possible. 4.Add or subtract terms on both sides of the equation to get all terms with the variable on one side of the equation. 5.Add or subtract a constant value on both sides of the equation to get all terms not containing the variable on the other side of the equation. 6.Divide both sides of the equation by the coefficient of the variable. 7.Simplify the solution, if possible. Procedure to Solve a Formula for a Specified Value 1.Remove any parentheses. 2.If fractions exist, multiply all terms on both sides by the LCD of the fractions. 3.Combine like terms, if possible. 4.Add or subtract terms on both sides of the equation to get all terms with the variable on one side of the equation. 5.Add or subtract a constant value on both sides of the equation to get all terms not containing the variable on the other side of the equation. 6.Divide both sides of the equation by the coefficient of the variable. 7.Simplify the solution, if possible.

10. 10 Solving a Formula for a Specified Value The formula for the area of a triangle is Example: If the area of a triangle is 66 inches, and the base is 8 inches, find the height of the triangle. b = 8 in h = ? A = 66 in 2 Multiply both sides by 2. Divide both sides by 8. The height of the triangle is 16.5 inches. Substitute in the values.

11. 11 Solving a Formula for a Specified Value The formula for the area of a trapezoid is Example: Solve the formula for b. Multiply both sides by 2. Divide both sides by h. Subtract a from both sides. The formula is now solved for b.

12. Graphing Linear Equations

13. 13 Graphing by Plotting Points The graph of any linear equation in two variables is a straight line. To Graph a Linear Equation 1.Look for three ordered pairs that are solutions to the equation. 2.Plot the points. 3.Draw a line through the points. To Graph a Linear Equation 1.Look for three ordered pairs that are solutions to the equation. 2.Plot the points. 3.Draw a line through the points.

14. 14 Graphing by Plotting Points xy 12 21 11 4 Example: Graph the equation y = –x + 3 by plotting points. (1, 2) is one of the solutions. Choose any value for x and find the corresponding y-value. For x = 1, y = –(1) + 3 = 2. Find three ordered pairs (solutions) to graph the line. x y 11 22 33 44 123 4 22 33 44 1 2 3 4 (  1, 4) (2, 1) (1, 2)

15. 15 Graphing by Plotting Intercepts The x-intercept of a line is the point where the line crosses the x-axis; it has the form (a, 0). The y-intercept of a line is the point where the line crosses the y-axis; it has the form (0, b). Intercept Method of Graphing 1.Find the x-intercept by letting y = 0 and solving for x. 2.Find the y-intercept by letting x = 0 and solving for y. 3.Find one additional ordered pair so that we have three points with which to plot the line. Intercept Method of Graphing 1.Find the x-intercept by letting y = 0 and solving for x. 2.Find the y-intercept by letting x = 0 and solving for y. 3.Find one additional ordered pair so that we have three points with which to plot the line.

16. 16 Graphing by Plotting Intercepts Let y = 0. x = 3 Example: Graph the equation –3y – 2x = – 6 by using the x- and y-intercepts. Let x = 0. –3y – 2(0) = – 6 y = 2 –3(0) – 2x = – 6 (3, 0) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 (0, 2) y-intercept x-intercept (  3, 4)

17. 17 Graphing Horizontal Lines y = 2 (0, 2) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 The graph of the equation y = b, where b is any real number, is a horizontal line through the point (0, b). The equation of the horizontal line. y =  3 (0,  3) x y 11 22 33 44 123 4 22 33 44 1 2 3 4

18. 18 Graphing Vertical Lines The graph of the equation x = a, where a is any real number, is a vertical line through the point (a, 0). x =  4 (  4, 0) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 The equation of the vertical line. x = 2 (2, 0) x y 11 22 33 44 123 4 22 33 44 1 2 3 4

19. The Slope of a Line

20. 20 Slope of a Line In a coordinate plane, the slope of a straight line is defined by the change in y divided by the change in x. (3, 0) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 (0, 2) Change in y =  2 Change in x = 3

21. 21 Definition of Slope of a Line The slope of any nonvertical line that contains the points with coordinates (x 1, y 1 ) and (x 2, y 2 ) is defined by the difference ratio Definition of Slope of a Line The slope of any nonvertical line that contains the points with coordinates (x 1, y 1 ) and (x 2, y 2 ) is defined by the difference ratio Slope of a Line where x 2  x 1. Example: Find the slope of the line that passes through (3, 6) and (1, 2).

22. 22 Slope of a Straight Line Slope of a Line Positive Slope Line goes up to the right x y 1. Lines with positive slopes go upward as x increases. Negative Slope Line goes downward to the right x y 2.Lines with negative slopes go downward as x increases. Continued.

23. 23 Slope of a Straight Line (continued) Slope of a Line Zero Slope Horizontal Line x y 3.Horizontal lines have a slope of 0. Undefined Slope Vertical Line x y 4.A vertical line has an undefined slope.

24. 24 Slope-Intercept Form of a Line The slope-intercept form of the equation of a line that has slope m and y-intercept (0, b) is given by y = mx + b. Slope-Intercept Form of a Line The slope-intercept form of the equation of a line that has slope m and y-intercept (0, b) is given by y = mx + b. Slope-Intercept Form of a Line y = 3x – 4 The slope is 3. The y-intercept is (0, -4). The slope is The y-intercept is

25. 25 Slope-Intercept Form of a Line Example: Find the equation of the line with slope and y-intercept (0,  5). slope: y-intercept:

26. 26 Graphing with Slope and y-Intercept Example: Graph the line with slope and y-intercept (0, 2). x y 11 22 33 44 123 4 22 33 44 1 2 3 4 rise run This is the graph of the line with slope and y-intercept (0, 2). 3 units down 2 units to the right (2,  1) (0, 2)

27. 27 Parallel Lines Parallel lines are two straight lines that never touch. Parallel lines have the same slope but different y-intercepts. Parallel Lines Parallel lines are two straight lines that never touch. Parallel lines have the same slope but different y-intercepts. m 1 = m 2 x y Slope m 1 Slope m 2

28. 28 Parallel Lines Example: Line c has a slope of If line d is parallel to line c, what is its slope? Parallel lines have the same slope. x y line d line c Line d has a slope of

29. 29 Perpendicular Lines Perpendicular lines are two lines that meet in a 90° angle. Perpendicular lines have a slope whose product is  1. If m 1 and m 2 are slopes of perpendicular lines, then Perpendicular Lines Perpendicular lines are two lines that meet in a 90° angle. Perpendicular lines have a slope whose product is  1. If m 1 and m 2 are slopes of perpendicular lines, then m 1 m 2 =  1 x y Slope m 1 Slope m 2 or m 1 =

30. 30 Perpendicular Lines Example: Line c has a slope of If line e is perpendicular to line c, what is its slope? Perpendicular lines have slopes whose product is  1. x y line e line c Line e has a slope of

31. Writing the Equation of a Line

32. 32 To Find the Equation of a Line Given a Point and the Slope 1.Substitute the given values of x, y, and m into the equation y = mx + b. 2.Solve for b. 3.Use the values of b and m to write the equation in the form y = mx + b. To Find the Equation of a Line Given a Point and the Slope 1.Substitute the given values of x, y, and m into the equation y = mx + b. 2.Solve for b. 3.Use the values of b and m to write the equation in the form y = mx + b. Finding the Equation Given a Point and the Slope

33. 33 Example: Find an equation of the line that passes through (4,  3) with a slope of 5. m = 5, x = 4, y =  3 The equation of the line is y = 5x  23. y = mx + b  3 = (5)(4) + b  23 = b Substitute known values. Finding the Equation Given a Point and the Slope

34. 34 Finding the Equation Given Two Points Example: Find an equation of the line that passes through (  2, 1) and (7, 4). Find the slope of the line.y = mx + b The equation of the line is

35. 35 Finding the Equation Given a Graph of the Line Example: Find the equation of the line for the following graph. Find the y-intercept. The equation of the line is (3, 0) x y 11 22 33 44 123 4 22 33 44 1 2 3 4 (0, 2) Change in y =  2 Change in x = 3 Find the slope

36. Systems of Linear Equations in Two Variables

37. 37 Systems of Equations A system of equations or system of inequalities is two or more equations or inequalities in several variables that are considered simultaneously. x y 11 22 33 123 22 33 1 2 3 The lines may intersect. x y 11 22 33 123 22 33 1 2 3 The lines may be parallel. x y 11 22 33 123 22 33 1 2 3 The lines may coincide.

38. 38 Systems of Equations A solution to a system of two linear equations in two variables is an ordered pair. Example: Determine if (  4, 16) is a solution to the system of equations. y =  4x 16 =  4(  4) 16 = 16 y =  2x + 8 16 =  2(  4) + 8 16 = 8 + 8 16 = 16 Yes, it is a solution. y =  2x + 8

39. 39 Graphing to Solve a System xy 22 0 0  6 1  9 Example: Solve by graphing. y =  3x  6 y = 2x  1 xy 22  5 0  1 23 x y 22 44 66 88 246 8 44 66 88 2 4 6 8 Continued.

40. 40 Graphing to Solve a System Example continued: x y 22 44 66 88 246 8 44 66 88 2 4 6 8 The lines intersect at (  1,  3). Check: y =  3x  6 y = 2x  1  3 =  3(  1)  6  3 = 3  6  3 =  3  3 = 2(  1)  1  3 =  2  1  3 =  3 y =  3x  6 y = 2x  1 A system of equations that has one solution is said to be consistent.

41. 41 Inconsistent Systems Example: Solve by graphing. 3x  2y =  4  9x + 6y =  1 x y 22 44 66 88 246 8 44 66 88 2 4 6 8 3x  2y =  4  9x + 6y =  1 A system of linear equations that has no solution is called an inconsistent system. The lines are parallel.

42. 42 Dependent Systems Example: Solve by graphing. 4x  6y = 8  2x + 3y =  4 x y 22 44 66 88 246 8 44 66 88 2 4 6 8 4x  6y = 8  2x + 3y =  4 A system of linear equations that has an infinite number of solutions is called a dependent system. The lines coincide.

43. 43 The Substitution Method How to Solve a System of Two Linear Equations by the Substitution Method 1.Choose one of the two equations and solve for one variable in terms of the other variable. 2.Substitute the expression from step1 into the other equation. 3.You now have one equation with one variable. Solve this equation for the variable. 4.Substitute this value for the variable into one of the original equations to obtain a value for the second variable. 5.Check the solution in both original equations.

44. 44 The Substitution Method Example: Find the solution. 2x – y = 13 – 4x – 9y = 7 y = 2x – 13 Solve the first equation for y. – 4x – 9y = 7 This is the original second equation. – 4x – 9(2x – 13) = 7 – 4x – 18x + 117 = 7 – 22x = – 110 x = 5 This variable is the easiest to isolate. Substitute the expression into the other equation and solve. Continued. Substitute this value into one of the original equations.

45. 45 The Substitution Method 2x – y = 13 – 4x – 9y = 7 First equation Substitute x = 5 into the first equation. Second equation 2x – y = 13 2(5) – y = 13 Solve for y.10 – y = 13 – y = 3 y = – 3The solution is (5, – 3). Continued. Example continued:

46. 46 The Substitution Method Check the solution (5, – 3) in both original equations. 2x – y = 13 – 4x – 9y = 7 2(5) – (  3) = 13 10 + 3 = 13 13 = 13 – 4(5) – 9(  3) = 7 – 20 + 27 = 7 7 = 7 Example continued:

47. 47 The Addition Method How to Solve a System of Two Linear Equations by the Addition (Elimination) Method 1.Arrange each equation in the form ax + by = c. (Remember that a, b, and c can be any real number.) 2.Multiply one or both equations by the appropriate numbers so that the coefficients of one of the variables are opposites. 3.Add the two equations from step 2 so that one variable is eliminated. 4.Solve the resulting equation for the remaining variable. 5.Substitute this value into one of the original equations to find the value of the other variable. 6.Check the solution in both of the original equations.

48. 48 The Addition Method Example: Solve by addition. 5x – 3y = 14 2x – y = 6 Multiply each term of equation (2) by  3.  x =  4 (  3)2x – (  3)y = (  3)6 Add the two equations. This equation is equivalent to equation (2).  6x + 3y =  18 5x – 3y = 14  6x + 3y =  18 x = 4 Substitute this value into either equation to find y. Continued.

49. 49 The Addition Method 5x – 3y = 14 2x – y = 6 8  y = 6 Substitute. 2(4) – y = 6  y =  2 The solution is (4, 2). Be sure to check the solution in both original equations. Example continued: 2x – y = 6 y = 2

50. 50 Identifying Inconsistent Systems Example: Solve the system algebraically. 3x + 6y = 12 x + 2y = 7 x =  2y + 7 Solve equation (2) for x. 3(  2y + 7) + 6y = 12 Substitute into equation (1).  6y + 21 + 6y = 12 Simplify. 21 = 12 This results in a false statement. There is no solution to this system of equations. If graphed, these lines would be parallel.

51. 51 Identifying Dependent Systems Example: Solve the system algebraically. (6)6x – (6)4y = (6)8 Multiply each term in (1) by 6. There are an infinite number of solutions to this system of equations. If graphed, these lines would be the same. 6x – 4y = 8 – 9x + 6y = –12 4(–9x) + 4(6y) = 4(–12) Multiply each term in (2) by 4. 36x – 24y = 48 This equation is equivalent to (1). –36x + 24y = –48 This equation is equivalent to (2). 0 = 0 Add the equations.

52. 52 Choosing an Appropriate Method Often becomes difficult to use if no variable has a coefficient of 1 or  1. Works well if one or more variable has a coefficient of 1 or  1. Substitution Works well if equations have fractional or decimal coefficients, or if no variable has a coefficient of 1 or  1. NoneAddition Advantage Disadvantage Method

53. 53 Choosing an Appropriate Method Example: Select an appropriate method for solving the system. 3x + 6y = 12 x + 2y = 7 b.) 6x – 4y = 8 – 9x + 6y = –12 c.) 5x – 3y = 13 0.9x + 0.4y = 30 a.) The substitution method should be used since x has a coefficient of 1. The addition method should be used since none of the variables have a coefficient of 1 or  1.

54. 54 Possible Solutions You obtain an equation that is always true. These equations are dependent. Infinite number of solutions You obtain an equation that is inconsistent with known facts. The system is inconsistent. No solution You obtain one value for x and one value for y. For example, x = 3, y = 5. One unique solution Graph Algebraic Interpretation Number of Solutions Two lines intersect at one point. Parallel lines Lines coincide (3, 5)