Illustrative Math Alignment: Grade 8 Unit 3

Topic

Linear Functions

Topic

Linear Functions

Topic

Linear Functions

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a horizontal line passing through the points (2, 3) and (7, 3). The slope is calculated as 0 since both y-coordinates are the same. Using the point-slope form, y - y1 = m(x - x1), the equation becomes y - 3 = 0(x - any x-value), which simplifies to y = 3, representing a horizontal line.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a horizontal line passing through the points (2, 3) and (7, 3). The slope is calculated as 0 since both y-coordinates are the same. Using the point-slope form, y - y1 = m(x - x1), the equation becomes y - 3 = 0(x - any x-value), which simplifies to y = 3, representing a horizontal line.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a horizontal line passing through the points (2, 3) and (7, 3). The slope is calculated as 0 since both y-coordinates are the same. Using the point-slope form, y - y1 = m(x - x1), the equation becomes y - 3 = 0(x - any x-value), which simplifies to y = 3, representing a horizontal line.

Topic

Linear Functions

Description

This example illustrates how to find the equation of a vertical line passing through the points (5, 3) and (5, 8). The slope is undefined because both x-coordinates are the same, resulting in division by zero when using the slope formula. This indicates a vertical line, and the equation is simply x = 5.

Topic

Linear Functions

Description

This example illustrates how to find the equation of a vertical line passing through the points (5, 3) and (5, 8). The slope is undefined because both x-coordinates are the same, resulting in division by zero when using the slope formula. This indicates a vertical line, and the equation is simply x = 5.

Topic

Linear Functions

Description

This example illustrates how to find the equation of a vertical line passing through the points (5, 3) and (5, 8). The slope is undefined because both x-coordinates are the same, resulting in division by zero when using the slope formula. This indicates a vertical line, and the equation is simply x = 5.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through the points (-6, 1) and (-2, 3). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 1/2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived and simplified to y = (1/2)x + 4.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through the points (-6, 1) and (-2, 3). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 1/2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived and simplified to y = (1/2)x + 4.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through the points (-6, 1) and (-2, 3). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 1/2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived and simplified to y = (1/2)x + 4.

Topic

Linear Functions

Description

This image shows a graph with two points (-8, 4) and (-4, 2). The slope is calculated as (y2 - y1) / (x2 - x1), resulting in a slope of -1/2. The equation of the line is derived using point-slope form and simplified to y = -(1/2)x. The slope is calculated as -1/2, and the line equation is determined using point-slope form: y = -(1/2)x.

Topic

Linear Functions

Description

This image shows a graph with two points (-8, 4) and (-4, 2). The slope is calculated as (y2 - y1) / (x2 - x1), resulting in a slope of -1/2. The equation of the line is derived using point-slope form and simplified to y = -(1/2)x. The slope is calculated as -1/2, and the line equation is determined using point-slope form: y = -(1/2)x.

Topic

Linear Functions

Description

This image shows a graph with two points (-8, 4) and (-4, 2). The slope is calculated as (y2 - y1) / (x2 - x1), resulting in a slope of -1/2. The equation of the line is derived using point-slope form and simplified to y = -(1/2)x. The slope is calculated as -1/2, and the line equation is determined using point-slope form: y = -(1/2)x.

Topic

Linear Functions

Description

This image shows a graph with two points (-7, 5) and (-1, 5). The slope is calculated as zero since the y-values are equal. The equation of the line is horizontal, simplified to y = 5. The slope is 0, indicating a horizontal line. The equation is y = 5.

Topic

Linear Functions

Description

This image shows a graph with two points (-7, 5) and (-1, 5). The slope is calculated as zero since the y-values are equal. The equation of the line is horizontal, simplified to y = 5. The slope is 0, indicating a horizontal line. The equation is y = 5.

Topic

Linear Functions

Description

This image shows a graph with two points (-7, 5) and (-1, 5). The slope is calculated as zero since the y-values are equal. The equation of the line is horizontal, simplified to y = 5. The slope is 0, indicating a horizontal line. The equation is y = 5.

Topic

Linear Functions

Description

This image shows a graph with two points (-5, 6) and (-5, 3). The slope is undefined because the x-values are equal. This results in a vertical line at x = -5. The slope is undefined, indicating a vertical line at x = -5.

Topic

Linear Functions

Description

This image shows a graph with two points (-5, 6) and (-5, 3). The slope is undefined because the x-values are equal. This results in a vertical line at x = -5. The slope is undefined, indicating a vertical line at x = -5.

Topic

Linear Functions

Description

This image shows a graph with two points (-5, 6) and (-5, 3). The slope is undefined because the x-values are equal. This results in a vertical line at x = -5. The slope is undefined, indicating a vertical line at x = -5.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -4) and (-4, -1) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated, and the equation is derived using point-slope form. The slope is calculated as (y2 - y1) / (x2 - x1) = (-1 - (-4)) / (-4 - (-5)) = 3 / 1 = 3. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 1 = 3(x + 4), which simplifies to y = 3x + 11.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -4) and (-4, -1) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated, and the equation is derived using point-slope form. The slope is calculated as (y2 - y1) / (x2 - x1) = (-1 - (-4)) / (-4 - (-5)) = 3 / 1 = 3. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 1 = 3(x + 4), which simplifies to y = 3x + 11.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -4) and (-4, -1) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated, and the equation is derived using point-slope form. The slope is calculated as (y2 - y1) / (x2 - x1) = (-1 - (-4)) / (-4 - (-5)) = 3 / 1 = 3. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 1 = 3(x + 4), which simplifies to y = 3x + 11.

Topic

The Point-Slope Form

Description

This example demonstrates how to find the equation of a line using the point-slope form. The given information includes a slope of 5 and a point (6, 5) through which the line passes. Using the point-slope formula y - y1 = m(x - x1), we can substitute the known values to derive the equation y - 5 = 5(x - 6). Simplifying this equation leads to the final result: y = 5x - 25.

Topic

The Point-Slope Form

Description

This example demonstrates how to find the equation of a line using the point-slope form. The given information includes a slope of 5 and a point (6, 5) through which the line passes. Using the point-slope formula y - y1 = m(x - x1), we can substitute the known values to derive the equation y - 5 = 5(x - 6). Simplifying this equation leads to the final result: y = 5x - 25.

Topic

The Point-Slope Form

Description

This example demonstrates how to find the equation of a line using the point-slope form. The given information includes a slope of 5 and a point (6, 5) through which the line passes. Using the point-slope formula y - y1 = m(x - x1), we can substitute the known values to derive the equation y - 5 = 5(x - 6). Simplifying this equation leads to the final result: y = 5x - 25.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -1 passing through the point (5, 2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 2 = -(x - 5). After simplification, the final equation of the line is y = -x + 7.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -1 passing through the point (5, 2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 2 = -(x - 5). After simplification, the final equation of the line is y = -x + 7.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -1 passing through the point (5, 2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 2 = -(x - 5). After simplification, the final equation of the line is y = -x + 7.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 3 passing through the point (-2, 7). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - 7 = 3(x - (-2)). After simplification, the resulting equation is y = 3x + 13.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 3 passing through the point (-2, 7). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - 7 = 3(x - (-2)). After simplification, the resulting equation is y = 3x + 13.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 3 passing through the point (-2, 7). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - 7 = 3(x - (-2)). After simplification, the resulting equation is y = 3x + 13.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -5 that passes through the point (-7, 5). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 5 = -5(x - (-7)). After simplification, the final equation of the line is y = -5x - 30.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -5 that passes through the point (-7, 5). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 5 = -5(x - (-7)). After simplification, the final equation of the line is y = -5x - 30.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of -5 that passes through the point (-7, 5). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - 5 = -5(x - (-7)). After simplification, the final equation of the line is y = -5x - 30.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 2 passing through the point (-2, -3). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-3) = 2(x - (-2)). After simplification, the resulting equation is y = 2x + 1.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 2 passing through the point (-2, -3). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-3) = 2(x - (-2)). After simplification, the resulting equation is y = 2x + 1.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of 2 passing through the point (-2, -3). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-3) = 2(x - (-2)). After simplification, the resulting equation is y = 2x + 1.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -7 passing through the point (-8, -5). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-5) = -7(x - (-8)). After simplification, the resulting equation is y = -7x - 61.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -7 passing through the point (-8, -5). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-5) = -7(x - (-8)). After simplification, the resulting equation is y = -7x - 61.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -7 passing through the point (-8, -5). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-5) = -7(x - (-8)). After simplification, the resulting equation is y = -7x - 61.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of 4 passing through the point (8, -2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - (-2) = 4(x - 8). After simplification, the final equation of the line is y = 4x - 34.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of 4 passing through the point (8, -2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - (-2) = 4(x - 8). After simplification, the final equation of the line is y = 4x - 34.

Topic

The Point-Slope Form

Description

In this example, we explore finding the equation of a line with a slope of 4 passing through the point (8, -2). Applying the point-slope formula y - y1 = m(x - x1), we substitute the given values to obtain y - (-2) = 4(x - 8). After simplification, the final equation of the line is y = 4x - 34.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -2 passing through the point (3, -8). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-8) = -2(x - 3). After simplification, the resulting equation is y = -2x - 2.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -2 passing through the point (3, -8). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-8) = -2(x - 3). After simplification, the resulting equation is y = -2x - 2.

Topic

The Point-Slope Form

Description

This example demonstrates the application of the point-slope formula to find the equation of a line with a slope of -2 passing through the point (3, -8). Using the formula y - y1 = m(x - x1), we substitute the given values to get y - (-8) = -2(x - 3). After simplification, the resulting equation is y = -2x - 2.

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates how to solve a proportion problem where two ratios a:b and c:d are proportional. Given the values b = 3, c = 4, and d = 6, we need to find the value of a. The proportion is set up as a / 3 = 4 / 6, which is then solved to find that a = 2.

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates how to solve a proportion problem where two ratios a:b and c:d are proportional. Given the values b = 3, c = 4, and d = 6, we need to find the value of a. The proportion is set up as a / 3 = 4 / 6, which is then solved to find that a = 2.