Illustrative Math-Media4Math Alignment

 

 

Illustrative Math Alignment: Grade 8 Unit 3

Linear Relationships

Lesson 6: More Linear Relationships

Use the following Media4Math resources with this Illustrative Math lesson.

Thumbnail Image Title Body Curriculum Topic
Math Example--Coordinate Geometry--Slope Formula: Example 4 Math Example--Coordinate Geometry--Slope Formula: Example 4 Math Example--Coordinate Geometry--Slope Formula: Example 4

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a vertical line passing through points (5, 8) and (5, 2) on a coordinate grid. When we apply the slope formula, we find that the slope is (8 - 2) / (5 - 5) = 6 / 0, which is undefined.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is vertical, resulting in an undefined slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 4 Math Example--Coordinate Geometry--Slope Formula: Example 4 Math Example--Coordinate Geometry--Slope Formula: Example 4

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a vertical line passing through points (5, 8) and (5, 2) on a coordinate grid. When we apply the slope formula, we find that the slope is (8 - 2) / (5 - 5) = 6 / 0, which is undefined.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is vertical, resulting in an undefined slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5 Math Example--Coordinate Geometry--Slope Formula: Example 5

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points in different quadrants: (-6, -2) in Quadrant III and (6, 5) in Quadrant I. Applying the slope formula, we find that the slope is (5 - (-2)) / (6 - (-6)) = 7 / 12 = 1 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-4, 8) in Quadrant II and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (8 - 2) / (-4 - 6) = 6 / -10 = -3 / 5.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example demonstrates how to handle points in different quadrants and interpret a negative slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-4, 8) in Quadrant II and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (8 - 2) / (-4 - 6) = 6 / -10 = -3 / 5.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example demonstrates how to handle points in different quadrants and interpret a negative slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6 Math Example--Coordinate Geometry--Slope Formula: Example 6

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-4, 8) in Quadrant II and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (8 - 2) / (-4 - 6) = 6 / -10 = -3 / 5.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example demonstrates how to handle points in different quadrants and interpret a negative slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a horizontal line connecting two points: (-4, 3) in Quadrant II and (2, 3) in Quadrant I. When we apply the slope formula, we find that the slope is (3 - 3) / (-4 - 2) = 0 / -6 = 0.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is horizontal, resulting in a slope of zero, even when the points are in different quadrants.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a horizontal line connecting two points: (-4, 3) in Quadrant II and (2, 3) in Quadrant I. When we apply the slope formula, we find that the slope is (3 - 3) / (-4 - 2) = 0 / -6 = 0.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is horizontal, resulting in a slope of zero, even when the points are in different quadrants.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7 Math Example--Coordinate Geometry--Slope Formula: Example 7

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a horizontal line connecting two points: (-4, 3) in Quadrant II and (2, 3) in Quadrant I. When we apply the slope formula, we find that the slope is (3 - 3) / (-4 - 2) = 0 / -6 = 0.

The slope formula is a key concept in coordinate geometry, helping us understand the steepness and direction of lines. This particular example highlights a special case where the line is horizontal, resulting in a slope of zero, even when the points are in different quadrants.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-2, -8) in Quadrant III and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (2 - (-8)) / (6 - (-2)) = 10 / 8 = 5 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-2, -8) in Quadrant III and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (2 - (-8)) / (6 - (-2)) = 10 / 8 = 5 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8 Math Example--Coordinate Geometry--Slope Formula: Example 8

Topic

Slope Formula

Description

This example illustrates the calculation of slope for a line connecting two points in different quadrants: (-2, -8) in Quadrant III and (6, 2) in Quadrant I. Applying the slope formula, we find that the slope is (2 - (-8)) / (6 - (-2)) = 10 / 8 = 5 / 4.

The slope formula is a crucial concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle negative coordinates and points in different quadrants when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points: (9, 6) and (2, -8) on a Cartesian plane. Applying the slope formula, we find that the slope is (6 - (-8)) / (9 - 2) = 14 / 7 = 2.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle points with both positive and negative coordinates when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points: (9, 6) and (2, -8) on a Cartesian plane. Applying the slope formula, we find that the slope is (6 - (-8)) / (9 - 2) = 14 / 7 = 2.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle points with both positive and negative coordinates when calculating slope.

Slope
Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9 Math Example--Coordinate Geometry--Slope Formula: Example 9

Topic

Slope Formula

Description

This example demonstrates the calculation of slope for a line connecting two points: (9, 6) and (2, -8) on a Cartesian plane. Applying the slope formula, we find that the slope is (6 - (-8)) / (9 - 2) = 14 / 7 = 2.

The slope formula is a fundamental concept in coordinate geometry, helping us understand the steepness and direction of lines. This example shows how to handle points with both positive and negative coordinates when calculating slope.

Slope
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 1 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 1 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 1

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + 1, where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 10 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 10 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 10

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 9x + (-2), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 2 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 2 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 2

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 6x + (-10), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 3 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 3 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 3

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -x + (-8), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 4 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 4 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 4

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 5 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 5 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 5

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -6x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 6 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 6 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 6

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -3x + 1, where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 7 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 7 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 7

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = -10x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 8 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 8 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 8

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + (-9), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 9 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 9 Math Example--Linear Function Concepts--Evaluating Linear Functions: Example 9

Topic

Linear Functions

Description

This example demonstrates how to evaluate a linear function at a specific value. The function in this example is the function f(x) = 7x + (-3), where x is replaced with a given value to determine the corresponding f(x) value. The process includes substitution and arithmetic simplification.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 1 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 1 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 1

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 2 and a y-intercept of 3. The process involves three key steps: first, plotting the y-intercept at (0, 3); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 10 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 10 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 10

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -4 and a y-intercept of 0. The method involves three main steps: plotting the y-intercept at the origin (0, 0), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 11 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 11 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 11

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of 0 and a y-intercept of 5. The procedure involves three key steps: plotting the y-intercept at (0, 5), recognizing that a slope of 0 results in a horizontal line, and drawing the line parallel to the x-axis through the y-intercept.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 12 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 12 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 12

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0 and a y-intercept of -5. The method involves three main steps: plotting the y-intercept at (0, -5), recognizing that a slope of 0 results in a horizontal line, and drawing the line parallel to the x-axis through the y-intercept.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 13 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 13 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 13

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 0 and a y-intercept of 0. The process involves recognizing that this special case results in a horizontal line coinciding with the x-axis. The line passes through the origin (0, 0) and extends infinitely in both directions along the x-axis.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 2 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 2 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 2

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0.5 and a y-intercept of 3. The method involves three main steps: plotting the y-intercept at (0, 3), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 3 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 3 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 3

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of 5 and a y-intercept of -4. The procedure involves three key steps: plotting the y-intercept at (0, -4), using the slope to determine a second point on the line, and connecting these points to form the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 4 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 4 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 4

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of 0.1 and a y-intercept of -4. The method involves three main steps: plotting the y-intercept at (0, -4), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 5 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 5 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 5

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of -4 and a y-intercept of 5. The process involves three key steps: first, plotting the y-intercept at (0, 5); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 6 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 6 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 6

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -1/3 and a y-intercept of 5. The method involves three main steps: plotting the y-intercept at (0, 5), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 7 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 7 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 7

Topic

Linear Functions

Description

This example demonstrates the process of graphing a linear function with a slope of -3 and a y-intercept of -2. The procedure involves three key steps: plotting the y-intercept at (0, -2), using the slope to determine a second point on the line, and connecting these points to form the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 8 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 8 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 8

Topic

Linear Functions

Description

This example illustrates the process of graphing a linear function with a slope of -0.25 and a y-intercept of -2. The method involves three main steps: plotting the y-intercept at (0, -2), using the slope to determine a second point on the line, and connecting these points to create the linear graph.

Slope-Intercept Form
Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 9 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 9 Math Example--Linear Function Concepts--Graphs of Linear Functions in Slope-Intercept Form: Example 9

Topic

Linear Functions

Description

This example demonstrates how to graph a linear function with a slope of 0.25 and a y-intercept of 0. The process involves three key steps: first, plotting the y-intercept at the origin (0, 0); second, using the slope to find another point on the line; and finally, connecting these points to form the line.

Slope-Intercept Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 1

Topic

Linear Functions

Description

This example demonstrates the process of converting a linear equation from standard form to slope-intercept form. The equation 2x + 4y = 8 is solved step-by-step, isolating y and dividing by its coefficient. The result is y = -1/2 x + 2, clearly showing the slope and y-intercept of the line.

Standard Form
Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 10 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 10 Math Example--Linear Function Concepts--Linear Equations in Standard Form: Example 10

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation x + y = 1 from standard form to slope-intercept form. The process involves isolating y, resulting in y = -x + 1. This simple transformation clearly reveals the slope and y-intercept of the line.

Standard Form