Illustrative Math Alignment: Grade 7 Unit 7

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

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

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 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 (-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

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

Solving Equations

Description

This math example focuses on solving percent equations, specifically asking "What is 5% of 8?" The solution involves converting 5% to its decimal form, 0.05, and then multiplying it by 8 to get the result of 0.4. This straightforward approach demonstrates how to tackle basic percent calculations efficiently.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "What is 170% of 9.5?" The solution involves converting 170% to its decimal equivalent, 1.7, and then multiplying it by 9.5 to obtain the result of 16.15. This example combines a percentage greater than 100% with a decimal base number, further illustrating the versatility of the percent-to-decimal conversion method in complex scenarios.

Topic

Solving Equations

Description

This math example focuses on solving percent equations, specifically asking "What is 225.5% of 78?" The solution involves converting 225.5% to its decimal form, 2.255, and then multiplying it by 78 to arrive at the answer of 175.89. This example introduces a decimal percentage greater than 200% and a larger whole number as the base value, demonstrating the scalability and flexibility of the percent-to-decimal conversion method in complex scenarios.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "What is 400% of 92.8?" The solution involves converting 400% to its decimal equivalent, 4.0, and then multiplying it by 92.8 to obtain the result of 371.2. This example showcases how to handle percentages greater than 100% and their application to decimal numbers, illustrating the versatility of the percent-to-decimal conversion method in complex scenarios.

Topic

Solving Equations

Description

This math example focuses on solving percent equations by asking "5 is what percent of 9?" The solution involves setting up the equation 9 * (x / 100) = 5, then solving for x to get x = 5 * (100 / 9), which is approximately 55.56%. This example introduces a new type of percent problem where students must find the percentage given two known values.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "6 is what percent of 2.3?" The solution involves setting up the equation 2.3 * (x / 100) = 6, then solving for x to get x = 6 * (100 / 2.3), which is approximately 260.87%. This example introduces a scenario where the resulting percentage is greater than 100% and involves a decimal base number.

Topic

Solving Equations

Description

This math example focuses on solving percent equations by asking "9 is what percent of 38?" The solution involves setting up the equation 38 * (x / 100) = 9, then solving for x to get x = 9 * (100 / 38), which is approximately 23.68%. This example demonstrates how to calculate a percentage when the first number is smaller than the second, resulting in a percentage less than 100%.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "2 is what percent of 55.5?" The solution involves setting up the equation 55.5 * (x / 100) = 2, then solving for x to get x = 2 * (100 / 55.5), which is approximately 3.6036%. This example introduces a scenario where the resulting percentage is a small fraction, less than 5%, and involves a decimal base number.

Topic

Solving Equations

Description

This math example focuses on solving percent equations by asking "8 is what percent of 120?" The solution involves setting up the equation 120 * (x / 100) = 8, then solving for x to get x = 8 * (100 / 120), which is approximately 6.67%. This example demonstrates how to calculate a percentage when dealing with larger whole numbers, resulting in a percentage less than 10%.

Topic

Solving Equations

Description

This math example demonstrates solving percent equations by asking "3.5 is what percent of 350?" The solution involves setting up the equation 350 * (x / 100) = 3.5, then solving for x to get x = 3.5 * (100 / 350), which equals 1%. This example introduces a scenario where the resulting percentage is a whole number (1%) and involves a decimal number as the first value.

Topic

Solving Equations

Description

This math example focuses on solving percent equations by asking "12 is what percent of 8?" The solution involves setting up the equation 8 * (x / 100) = 12, then solving for x to get x = 12 * (100 / 8), which equals 150%. This example demonstrates how to calculate a percentage when the first number is larger than the second, resulting in a percentage greater than 100%.