1 / 36

Solving Systems of Equations 2-2: Linear Equations

Essential Question: How can you use the equations of two non-vertical lines to tell whether the equations are parallel or perpendicular?. Solving Systems of Equations 2-2: Linear Equations. 2-2: Linear Equations. Graphing a Linear Equation Any equation with both x and y can be graphed.

guido
Télécharger la présentation

Solving Systems of Equations 2-2: Linear Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Essential Question: How can you use the equations of two non-vertical lines to tell whether the equations are parallel or perpendicular? Solving Systems of Equations2-2: Linear Equations

  2. 2-2: Linear Equations • Graphing a Linear Equation • Any equation with both x and y can be graphed. • A solution of a linear equation is any ordered pair (x, y) that makes the equation true. • In the equation y = 3x + 2 if x = 4 then y = 3(4) + 2 = 14 • Which means (4, 14) is a solution to that equation. • There are infinite solutions to a linear equation, you just need to decide on an x value to start with. • Because y depends on the value of x, we call y the dependent variable, and call x the independent variable

  3. 2-2: Linear Equations • Graphing a Linear Equation • You can graph a linear equation by taking two solutions, putting them on a coordinate plane, and connecting a line through them. • Example: Graph the equation y = 2/3x + 3

  4. 2-2: Linear Equations • Graphing a Linear Equation • You can graph a linear equation by taking two solutions, putting them on a coordinate plane, and connecting a line through them. • Example: Graph the equation y = 2/3x + 3 • Let x = 3, then y = 2/3(3) + 3 = 5 • Use the point (3, 5)

  5. 2-2: Linear Equations • Graphing a Linear Equation • You can graph a linear equation by taking two solutions, putting them on a coordinate plane, and connecting a line through them. • Example: Graph the equation y = 2/3x + 3 • Let x = 3, then y = 2/3(3) + 3 = 5 • Use the point (3, 5) • Let x = -3, then y = 2/3(-3) + 3 = 1 • Use the point (-3,1)

  6. 2-2: Linear Equations • Graphing a Linear Equation • Use the point (3, 5) • Use the point (-3,1) Connect the line

  7. 2-2: Linear Equations • The y-intercept is where the graph crosses the y-axis. • It can be found by setting x = 0 in a linear equation. • The x-intercept is where the graph crosses the x-axis. • It can be found by setting y = 0 in a linear equation. • The intercepts can also be used in graphing a linear equation.

  8. 2-2: Linear Equations • Example: The equation 3x + 2y = 120models the number of passengers who can sit in a train car, where x is the number of adults and y is the number of children. • Graph the equation. Explain what the x- and y-intercepts represent.Describe the domain and range.

  9. 2-2: Linear Equations • 3x + 2y = 120 • x-intercept • 3x + 2(0) = 120 • 3x = 120 • x = 40

  10. 2-2: Linear Equations • 3x + 2y = 120 • x-intercept • 3x + 2(0) = 120 • 3x = 120 • x = 40 • y-intercept • 3(0) + 2y = 120 • 2y = 120 • y = 60

  11. 2-2: Linear Equations • 3x + 2y = 120 • x-intercept • 3x + 2(0) = 120 • 3x = 120 • x = 40 • y-intercept • 3(0) + 2y = 120 • 2y = 120 • y = 60 • Use the points (40,0) and (0,60)

  12. 2-2: Linear Equations • Slope • Slope is found by taking the vertical change and dividing by the horizontal change • Rise over Run • The formula is:where (x1,y1) and (x2,y2) represent solutions to a linear equation

  13. 2-2: Linear Equations • Slope • Example: Find the slope of the line through the points (3,2) and (-9,6)

  14. 2-2: Linear Equations • Writing Equations of Lines • Slope-Intercept Form • Used when you know the slope and the y-intercept • y = mx + b • What does the m stand for? What does the b stand for? • Example: Find the slope of 4x + 3y = 7

  15. 2-2: Linear Equations • Writing Equations of Lines • Slope-Intercept Form • Used when you know the slope and the y-intercept • y = mx + b • slopey-intercept • Once you get y by itself, the slope is the coefficient in front of the “x”. • Example: Find the slope of 4x + 3y = 7

  16. 2-2: Linear Equations • 4x + 3y = 7 Get y by itself

  17. 2-2: Linear Equations • 4x + 3y = 7-4x -4x • 3y = -4x + 7

  18. 2-2: Linear Equations • 4x + 3y = 7-4x -4x • 3y = -4x + 73 3 3 • y = -4/3 x + 7/3(leave as fractions)

  19. 2-2: Linear Equations • 4x + 3y = 7-4x -4x • 3y = -4x + 73 3 3 • y = -4/3 x + 7/3(leave as fractions) • Slope = -4/3

  20. 2-2: Linear Equations • 4x + 3y = 7-4x -4x • 3y = -4x + 73 3 3 • y = -4/3 x + 7/3(leave as fractions) • Slope = -4/3 • y-intercept = 7/3

  21. 2-2: Linear Equations • Assignment • Page 67 • 1 – 19, 33 – 37 (odd problems) • SHOW WORK

  22. Essential Question: How can you use the equations of two non-vertical lines to tell whether the equations are parallel or perpendicular? Solving Systems of Equations2-2: Linear Equations (Day 2)

  23. 2-2: Linear Equations • Writing Equations of Lines • Point-Slope Form • Used when you know a point on the line and the slope • y – y1 = m(x – x1) • Example: Write in slope-intercept form an equation of the line with slope -1/2 through the point (8,-1)

  24. 2-2: Linear Equations • Slope = -1/2 Point = (8, -1) • y – y1 = m(x – x1)

  25. 2-2: Linear Equations • Slope = -1/2 Point = (8, -1) • y – y1 = m(x – x1) (replace) • y – (-1) = -1/2(x – 8)

  26. 2-2: Linear Equations • Slope = -1/2 Point = (8, -1) • y – y1 = m(x – x1) (replace) • y – (-1) = -1/2(x – 8) (distribute) • y + 1 = -1/2x + 4

  27. 2-2: Linear Equations • Slope = -1/2 Point = (8, -1) • y – y1 = m(x – x1) (replace) • y – (-1) = -1/2(x – 8) (distribute) • y + 1 = -1/2x + 4 (subtract 1) • -1 -1 • y = -1/2 x + 3

  28. 2-2: Linear Equations • Writing Equations of Lines • Sometimes, you will be given two points. In this case, you will first need to find the slope of the two points, then use either one of the points and the slope in point-slope form • Example: Write in point-slope form an equation of the line through (1,5) and (4,-1)

  29. 2-2: Linear Equations • Writing Equations of Lines • Sometimes, you will be given two points. In this case, you will first need to find the slope of the two points, then use either one of the points and the slope in point-slope form • Example: Write in point-slope form an equation of the line through (1,5) and (4,-1) • Find the slope:

  30. 2-2: Linear Equations • Writing Equations of Lines • Example: Write in point-slope form an equation of the line through (1,5) and (4,-1) • Find the slope: • Choose either point • y – y1 = m(x – x1) • y – 5 = -2(x – 1) OR y + 1 = -2(x – 4)will give valid answers

  31. 2-2: Linear Equations • The slopes of horizontal, vertical, parallel and perpendicular lines have special properties

  32. 2-2: Linear Equations • Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2 • What is the slope of my starting line?

  33. 2-2: Linear Equations • Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2 • What is the slope of my starting line? ¾ • What is the slope of my perpendicular line?

  34. 2-2: Linear Equations • Write an equation of the line through the point (6,1) and perpendicular to y = ¾ x + 2 • What is the slope of my starting line? ¾ • What is the slope of my perpendicular line? • Flip fraction = 4/3 • Flip sign = -4/3 • Leave your answer in slope-intercept form • You have the slope, so find the intercept • Note: You can also solve using point-slope form

  35. 2-2: Linear Equations • Slope = -4/3 Point = (6, 1) • y = mx + b • Find b • 1 = -4/3 (6) + b (replace) • 1 = -8 + b • +8 +8 • 9 = b • y = -4/3 x + 9

  36. 2-2: Linear Equations • Assignment • Page 68 • #21,23, 25 (Write in slope-intercept form) • #27 – 31 (odd problems) • #38 & 39

More Related