Illustrative Math Alignment: Grade 8 Unit 3

Topic

Linear Functions

Description

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

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -x - y = -1 from standard form to slope-intercept form. The process involves manipulating the equation to solve for y, yielding y = -x + 1. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

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

Topic

Linear Functions

Description

This example demonstrates 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 transformation clearly reveals the slope and y-intercept of the line.

Linear functions are fundamental mathematical concepts that describe relationships between two variables. The examples in this collection, such as showing step-by-step transformations from standard form to slope-intercept form, help in understanding how each part of the equation affects the graph and the relationship itself.

Topic

Linear Functions

Description

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

Topic

Linear Functions

Description

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

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation 12x + 28y = 0 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = -3/7 x. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation 3x + 6y = -18 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = -1/2 x - 3. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -14x - 35y = 0 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = -2/5 x. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation 28x - 21y = 0 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = 4/3 x. This step-by-step solution clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the conversion of the linear equation -13x + 39y = 0 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = 1/3 x. This process clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the transformation of the linear equation 5x - 2y = 12 from standard form to slope-intercept form. The process involves isolating y and dividing by its coefficient, resulting in y = 2.5x - 6. This step-by-step solution clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -6x + 10y = 20 from standard form to slope-intercept form. The solution process involves isolating y and dividing by its coefficient, resulting in y = 0.6x + 2. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the process of converting the linear equation -9x - 15y = -45 from standard form to slope-intercept form. The solution involves rearranging the equation to isolate y, resulting in y = 3/5 x + 3. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the conversion of the linear equation -8x - 12y = 25 from standard form to slope-intercept form. The process involves manipulating the equation to solve for y, yielding y = -2/3 x - 25/12. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example illustrates the transformation of the linear equation -2x + 16y = -36 from standard form to slope-intercept form. The solution process involves solving for y, resulting in y = 1/8 x - 9/4. This step-by-step conversion clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example showcases the conversion of the linear equation -7x + 21y = 63 from standard form to slope-intercept form. The process involves rearranging the equation to isolate y, resulting in y = 1/3 x + 3. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

This example demonstrates the process of converting the linear equation 10x - 28y = 56 from standard form to slope-intercept form. The solution involves isolating y and dividing by its coefficient, resulting in y = (5/14)x - 2. This transformation clearly reveals the slope and y-intercept of the line.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 28x + 7y = 35 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -4x + 5 by isolating y. The y-intercept is found as 5 (where x = 0), and the x-intercept is calculated as 1.25 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 6x - 2y = 6 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 3x - 3. The y-intercept is -3 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 24x + 12y = 48 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -2x + 4. The y-intercept is 4 (where x = 0), and the x-intercept is 2 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 15x + 3y = 15 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -5x + 5. The y-intercept is 5 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 12x + 3y = 21 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -4x + 7. The y-intercept is 7 (where x = 0), and the x-intercept is 1.75 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 10x + 5y = 15 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = -2x + 3. The y-intercept is 3 (where x = 0), and the x-intercept is 1.5 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 10x - 2y = 10 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 5x - 5. The y-intercept is -5 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 32x - 8y = 32 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 4x - 4. The y-intercept is -4 (where x = 0), and the x-intercept is 1 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 16x - 8y = 40 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 2x - 5. The y-intercept is -5 (where x = 0), and the x-intercept is 2.5 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

Find the x- and y-intercepts for the linear equation 20x - 10y = 60 given in standard form. Convert it to slope-intercept form.

The equation is rewritten as y = 2x - 6. The y-intercept is -6 (where x = 0), and the x-intercept is 3 (where y = 0). These intercepts are plotted on a graph.

Linear functions are a fundamental concept in algebra, where we explore the relationships between variables. The examples in this collection highlight how to find intercepts and analyze the relationships they represent in a linear context.

Topic

Linear Functions

Description

This example demonstrates how to find the equation of a line passing through two given points: (6, 4) and (8, 8). The slope is calculated using the formula (y2 - y1) / (x2 - x1), resulting in a slope of 2. Using the point-slope form of a line, y - y1 = m(x - x1), the equation is derived as y - 8 = 2(x - 8), which simplifies to y = 2x - 8.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-6, -2) and (-2, -6) 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) = (-6 - (-2)) / (-2 - (-6)) = -4 / 4 = -1. Using point-slope form, y - y1 = m(x - x1), the equation is derived as y + 2 = -(x + 6), which simplifies to y = -x - 8.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-8, -4) and (-2, -4) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is calculated as zero, indicating a horizontal line. The slope is calculated as (y2 - y1) / (x2 - x1) = (-4 - (-4)) / (-8 - (-2)) = 0 / -6 = 0. Since the slope is zero, it indicates a horizontal line at y = -4. Using point-slope form, the equation becomes y + 4 = 0, which simplifies to y = -4.

Topic

Linear Functions

Description

The image shows a coordinate plane with two points (-5, -8) and (-5, -3) marked. It provides a step-by-step solution to find the equation of the line passing through these points. The slope is undefined, indicating a vertical line. The slope is calculated as (y2 - y1) / (x2 - x1) = (-3 - (-8)) / (-5 - (-5)) = 5 / 0, which is undefined. Since the slope is undefined, it indicates a vertical line at x = -5. Therefore, the equation of the line is simply x = -5.

Topic

Linear Functions

Description

This image shows a graph with two points (2, -4) and (6, -2) marked. The example demonstrates how to find the equation of a line passing through these points. The slope is calculated as 1/2, and the point-slope form is used to derive the equation of the line. The slope formula is used: (y2 - y1) / (x2 - x1) = (-2 - (-4)) / (6 - 2) = 2 / 4 = 1/2. Then, using the point-slope form: y - (-2) = (1/2)(x - 6), which simplifies to y = (1/2)x - 5.

Topic

Linear Functions

Description

This image shows a graph with two points (2, -3) and (5, -6). The example demonstrates how to find the equation of a line passing through these points. The slope is calculated as -1, and the point-slope form is used to derive the equation. The slope formula is used: (y2 - y1) / (x2 - x1) = (-3 - (-6)) / (2 - 5) = 3 / -3 = -1. Then, using the point-slope form: y - (-3) = -(x - 2), which simplifies to y = -x - 1.

Topic

Linear Functions

Description

This image shows a graph with two points (3, -4) and (7, -4). The example demonstrates how to find the equation of a horizontal line passing through these points. The slope is calculated as 0, and the point-slope form is used to derive the equation. The slope formula is used: (y2 - y1) / (x2 - x1) = (-4 - (-4)) / (7 - 3) = 0 / 4 = 0. Then, using the point-slope form: y - (-4) = 0(x - 3), which simplifies to y = -4.

Topic

Linear Functions

Description

This image shows a graph with two points (4, -8) and (4, -2). The example demonstrates how to find the equation of a vertical line passing through these points. The slope is undefined because division by zero occurs in calculating it. The slope formula is used: (y2 - y1) / (x2 - x1) = (-2 - (-8)) / (4 - 4) = 6 / 0 = undefined. Since this represents a vertical line, its equation is simply x = 4.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-2, 0.5) and (5, 4). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (4 - 0.5) / (5 - (-2)) = 3.5 / 7 = 1 / 2. The point-slope form is used to find the equation: y - 4 = (1 / 2)(x - 5), resulting in y = (1 / 2)x + 1 / 2.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-3, 5) and (5, 1). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (5 - 1) / (-3 - 5) = -1 / 2. The point-slope form is used to find the equation: y - 5 = (-1 / 2)(x + 3), resulting in y = (-1 / 2)x + 7 / 2.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-4, 3) and (6, 3). The example demonstrates how to find the equation of a horizontal line using the slope formula and point-slope form. The slope is calculated as (3 - 3) / (-4 - 6) = 0. Since the slope is zero, the equation of the line is simply y = 3, indicating a horizontal line passing through y = 3.

Topic

Linear Functions

Description

This example illustrates the process of finding the equation of a line passing through the points (3, 7) and (9, 1). The slope is calculated as -1 using the formula (y2 - y1) / (x2 - x1). Employing the point-slope form, y - y1 = m(x - x1), the equation is derived as y - 1 = -(x - 9), which simplifies to y = -x + 10.

Topic

Linear Functions

Description

This image shows a graph with two points plotted at (-3, -5) and (5, 1). The example demonstrates how to find the equation of a line using the slope formula and point-slope form. The slope is calculated as (1 - (-5)) / (5 - (-3)) = 6 / 8 = 3 / 4. The point-slope form is used to find the equation: y - 1 = (3 / 4)(x - 5), resulting in y = 3/4x - 2 3/4.

Topic

Linear Functions

Description

The image shows a graph with two points (2, -4) and (6, 1) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (1 - (-4)) / (6 - 2) = 5 / 4. Using point-slope form: y - 1 = (5 / 4)(x - 6), which simplifies to y = 5/4x - 6 1/2.

Topic

Linear Functions

Description

The image shows a graph with two points (5, 1) and (8, -5) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (1 - (-5)) / (5 - 8) = -2. Using point-slope form: y - 1 = -2(x - 5), which simplifies to y = -2x + 11.

Topic

Linear Functions

Description

The image shows a graph with two points (6, 1) and (6, -5) marked on a coordinate plane. The example demonstrates how to find the equation of a vertical line using these two points. Since both x-coordinates are equal, the slope is undefined, indicating a vertical line. The slope is undefined because x-values are equal: (1 - (-5)) / (6 - 6) = undefined. This means it's a vertical line at x = 6.

Topic

Linear Functions

Description

The image shows a graph with two points (-6, -2) and (-3, 4) marked on a coordinate plane. The example demonstrates how to find the equation of a line using these two points. The slope is calculated using the slope formula, and then the point-slope form is applied to derive the equation of the line. The slope is calculated as (4 - (-2)) / (-3 - (-6)) = 2. Using point-slope form: y - 4 = 2(x + 3), which simplifies to y = 2x + 10.

Topic

Linear Functions

Description

A graph with two points (-8, 6) and (-3, -4) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = -2x - 10. The slope is calculated as (-4 - 6) / (-3 - (-8)) = -2. Using the point-slope form y - y1 = m(x - x1), the equation becomes y - 6 = -2(x + 8), which simplifies to y = -2x - 10.

Topic

Linear Functions

Description

A graph with two points (-8, -6) and (4, 2) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = 1/4 x - 3. The slope is calculated as (2 + 6) / (4 + 8) = 1/4. Using the point-slope form y - y1 = m(x - x1), the equation becomes y + 2 = 1/4(x - 4), which simplifies to y = 1/4 x - 3.

Topic

Linear Functions

Description

A graph with two points (-3, 4) and (-3, -5) marked on a coordinate plane. The slope is undefined because both x-coordinates are equal, leading to division by zero. This represents a vertical line at x = -3. The slope is undefined because (4 + 5) / (-3 + 3) results in division by zero. This indicates a vertical line at x = -3.

Topic

Linear Functions

Description

A graph with two points (-6, -1) and (6, 4) marked on a coordinate plane. The slope is calculated using the formula (y2 - y1) / (x2 - x1), and the equation of the line is derived using the point-slope form. The final equation is y = -(1/4)x - 2.5 or -(1/4)x - (5/2). The slope is calculated as (4 + 1) / (6 + 6) = -(1/4). Using the point-slope form y - y1 = m(x - x1), the equation becomes y + 1 = -(1/4)(x + 6), which simplifies to y = -(1/4)x - (5/2).

Topic

Linear Functions