Illustrative Math-Media4Math Alignment

 

 

Illustrative Math Alignment: Grade 8 Unit 3

Linear Relationships

Lesson 5: Introduction to Linear Relationships

Use the following Media4Math resources with this Illustrative Math lesson.

Thumbnail Image Title Body Curriculum Topic
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 4

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 5

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 6

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 7

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8 Math Example--Linear Function Concepts--The Point-Slope Formula: Example 8

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.

 Point-Slope Form
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1

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.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 1

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.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with sides of 6 and 9, and the other with sides of 2x and 2x + 9. The problem requires setting up a proportion: 6 / 9 = 2x / (2x + 9). Solving this equation leads to x = 9, which then allows us to find the side lengths of 18 and 27.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 10

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with sides of 6 and 9, and the other with sides of 2x and 2x + 9. The problem requires setting up a proportion: 6 / 9 = 2x / (2x + 9). Solving this equation leads to x = 9, which then allows us to find the side lengths of 18 and 27.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, one with legs of 7 and 9, and the other with legs of 14 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 7 / 9 = 14 / x. Solving this equation leads to x = 18.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 11

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, one with legs of 7 and 9, and the other with legs of 14 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 7 / 9 = 14 / x. Solving this equation leads to x = 18.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with legs of 5 and 8, and the other with legs of 2x and (2x + 6). The problem requires setting up a proportion: 5 / 8 = 2x / (2x + 6). Solving this equation leads to x = 5, which then allows us to find the side lengths of 10 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 12

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with legs of 5 and 8, and the other with legs of 2x and (2x + 6). The problem requires setting up a proportion: 5 / 8 = 2x / (2x + 6). Solving this equation leads to x = 5, which then allows us to find the side lengths of 10 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles with special angles (30°-60°-90°). Two triangles are shown, with the smaller one having sides of 6√3 and 6, and the larger one having sides of x and 12. The problem requires finding the length of side x by setting up a proportion: (6√3) / 6 = x / 12. Solving this equation leads to x = 12√3.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 13

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles with special angles (30°-60°-90°). Two triangles are shown, with the smaller one having sides of 6√3 and 6, and the larger one having sides of x and 12. The problem requires finding the length of side x by setting up a proportion: (6√3) / 6 = x / 12. Solving this equation leads to x = 12√3.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with special angles (30°-60°-90°) and algebraic expressions. Two triangles are shown, with the smaller one having sides of 16 and 8, and the larger one having sides of (3x + 4) and 2x. The problem requires setting up a proportion: 16 / 8 = (3x + 4) / 2x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 8 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 14

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with special angles (30°-60°-90°) and algebraic expressions. Two triangles are shown, with the smaller one having sides of 16 and 8, and the larger one having sides of (3x + 4) and 2x. The problem requires setting up a proportion: 16 / 8 = (3x + 4) / 2x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 8 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, with the smaller one having sides of 4 and 3, and the larger one having sides of x and 12. The problem requires finding the length of side x by setting up a proportion: 4 / 3 = x / 12. Solving this equation leads to x = 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 15

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, with the smaller one having sides of 4 and 3, and the larger one having sides of x and 12. The problem requires finding the length of side x by setting up a proportion: 4 / 3 = x / 12. Solving this equation leads to x = 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, with the smaller one having sides of (3x + 4) and 6, and the larger one having sides of 3x and 3x. The problem requires setting up a proportion: 8 / 6 = (3x + 4) / 3x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 12 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 16

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, with the smaller one having sides of (3x + 4) and 6, and the larger one having sides of 3x and 3x. The problem requires setting up a proportion: 8 / 6 = (3x + 4) / 3x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 12 and 16.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, with the left one having sides of 12 and 5, and the right one having sides of x and 15. The problem requires finding the length of side x by setting up a proportion: 12 / 5 = x / 15. Solving this equation leads to x = 36.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 17

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar right triangles. Two right triangles are shown, with the left one having sides of 12 and 5, and the right one having sides of x and 15. The problem requires finding the length of side x by setting up a proportion: 12 / 5 = x / 15. Solving this equation leads to x = 36.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, with the left one having sides of 12 and 5, and the right one having sides of 10x + 8 and 5x. The problem requires setting up a proportion: 12 / 5 = (10x + 8) / 5x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 20 and 48.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 18

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, with the left one having sides of 12 and 5, and the right one having sides of 10x + 8 and 5x. The problem requires setting up a proportion: 12 / 5 = (10x + 8) / 5x. Solving this equation leads to x = 4, which then allows us to find the side lengths of 20 and 48.

 Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 19 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 19 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 19

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar isosceles right triangles. Two 45-45-90 triangles are shown, with one having sides labeled y and 5√2, and the other having sides labeled x and 12. The problem requires finding the lengths of both x and y using the special properties of isosceles right triangles and proportions.

 Proportions