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**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

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**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

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**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

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**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)

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**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)

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**2-2: Linear Equations Graphing a Linear Equation Use the point (3, 5)**

Connect the line

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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.

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2-2: Linear Equations Example: The equation 3x + 2y = 120 models 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.

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**2-2: Linear Equations 3x + 2y = 120 x-intercept 3x + 2(0) = 120**

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**2-2: Linear Equations 3x + 2y = 120 x-intercept y-intercept**

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**2-2: Linear Equations 3x + 2y = 120 x-intercept y-intercept**

Use the points (40,0) and (0,60)

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**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

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**2-2: Linear Equations Slope**

Example: Find the slope of the line through the points (3,2) and (-9,6)

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**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

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**2-2: Linear Equations Writing Equations of Lines Slope-Intercept Form**

Used when you know the slope and the y-intercept y = mx + b slope y-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

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

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

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**2-2: Linear Equations 4x + 3y = 7 -4x -4x 3y = -4x + 7 3 3 3**

y = -4/3 x + 7/3 (leave as fractions)

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**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

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**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

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**2-2: Linear Equations Assignment Page 67**

1 – 19, 33 – 37 (odd problems) SHOW WORK

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**Solving Systems of Equations 2-2: Linear Equations (Day 2)**

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 (Day 2)

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**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)

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

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**2-2: Linear Equations Slope = -1/2 Point = (8, -1)**

y – y1 = m(x – x1) (replace) y – (-1) = -1/2(x – 8)

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**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

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**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) y = -1/2 x + 3

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**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)

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**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:

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**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

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2-2: Linear Equations The slopes of horizontal, vertical, parallel and perpendicular lines have special properties Horizontal Line Vertical Line Parallel Lines Perpendicular Lines m = 0 m = undefined Slopes are equal Slopes are inverse reciprocals y is constant x is constant Flip the fraction Flip the sign

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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?

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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?

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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

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**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

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**2-2: Linear Equations Assignment Page 68**

#21,23, 25 (Write in slope-intercept form) #27 – 31 (odd problems) #38 & 39

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