Illustrative Math Alignment: Grade 7 Unit 2
Introducing Proportional Relationships
Lesson 10: Introducing Graphs of Proportional Relationships
Use the following Media4Math resources with this Illustrative Math lesson.
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar isosceles right triangles with algebraic expressions. Two 45-45-90 triangles are shown, with one having sides labeled 15√2 and 15, and the other having sides labeled √2 * 5x and y. The problem requires finding the length of y in terms of x using the special properties of isosceles right triangles and proportions. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar isosceles right triangles with algebraic expressions. Two 45-45-90 triangles are shown, with one having sides labeled 15√2 and 15, and the other having sides labeled √2 * 5x and y. The problem requires finding the length of y in terms of x using the special properties of isosceles right triangles and proportions. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar isosceles triangles. Two isosceles triangles are shown, with the smaller one having sides of 30 and 16, and the larger one having sides of x and 20. The problem requires finding the length of side x by setting up a proportion: 30 / 16 = x / 20. Solving this equation leads to x = 37.5. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar isosceles triangles. Two isosceles triangles are shown, with the smaller one having sides of 30 and 16, and the larger one having sides of x and 20. The problem requires finding the length of side x by setting up a proportion: 30 / 16 = x / 20. Solving this equation leads to x = 37.5. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar isosceles triangles with algebraic expressions. Two isosceles triangles are shown, with the smaller one having sides of 22 and 12, and the larger one having sides of 10x + 5 and 6x. The problem requires setting up a proportion: 22 / 12 = (10x + 5) / 6x. Solving this equation leads to x = 5, which then allows us to find the side lengths of the larger triangle. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar isosceles triangles with algebraic expressions. Two isosceles triangles are shown, with the smaller one having sides of 22 and 12, and the larger one having sides of 10x + 5 and 6x. The problem requires setting up a proportion: 22 / 12 = (10x + 5) / 6x. Solving this equation leads to x = 5, which then allows us to find the side lengths of the larger triangle. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar equilateral triangles. Two equilateral triangles are shown, with the smaller one having a side length of 12, and the larger one having a side length of 6x and an additional side labeled as 12 + x. The problem requires finding the lengths of the sides of the larger triangle by setting up a proportion: 12 / 12 = (12 + x) / 6x. Solving this equation leads to x = 2.4, and the length of the larger triangle's side is found to be approximately 14.4 units. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar equilateral triangles. Two equilateral triangles are shown, with the smaller one having a side length of 12, and the larger one having a side length of 6x and an additional side labeled as 12 + x. The problem requires finding the lengths of the sides of the larger triangle by setting up a proportion: 12 / 12 = (12 + x) / 6x. Solving this equation leads to x = 2.4, and the length of the larger triangle's side is found to be approximately 14.4 units. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a complex proportion problem using similar equilateral triangles with algebraic expressions. Two equilateral triangles are shown, with the smaller one having a side length of 3x, and the larger one having side lengths labeled as (4x + 1) and (x + 1). The problem requires setting up a proportion: 3x / (4x + 1) = 3x / (3x + (x + 1)). Solving this equation leads to a quadratic equation, which when solved gives x ≈ 7/1. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a complex proportion problem using similar equilateral triangles with algebraic expressions. Two equilateral triangles are shown, with the smaller one having a side length of 3x, and the larger one having side lengths labeled as (4x + 1) and (x + 1). The problem requires setting up a proportion: 3x / (4x + 1) = 3x / (3x + (x + 1)). Solving this equation leads to a quadratic equation, which when solved gives x ≈ 7/1. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar rectangles. Two rectangles are shown, with the smaller one having sides of 2 and 3, and the larger one having sides of x and 15. The problem requires finding the length of side x by setting up a proportion: 2 / 3 = x / 15. Solving this equation leads to x = 10. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar rectangles. Two rectangles are shown, with the smaller one having sides of 2 and 3, and the larger one having sides of x and 15. The problem requires finding the length of side x by setting up a proportion: 2 / 3 = x / 15. Solving this equation leads to x = 10. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with algebraic expressions. Two rectangles are shown, with the smaller one having sides of 5 and 12, and the larger one having sides of x + 5 and 3x + 6. The problem requires setting up a proportion: 5 / 12 = (x + 5) / (3x + 6). Solving this equation leads to x = 10, which then allows us to find the side lengths of the larger rectangle as 15 and 36. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with algebraic expressions. Two rectangles are shown, with the smaller one having sides of 5 and 12, and the larger one having sides of x + 5 and 3x + 6. The problem requires setting up a proportion: 5 / 12 = (x + 5) / (3x + 6). Solving this equation leads to x = 10, which then allows us to find the side lengths of the larger rectangle as 15 and 36. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms. Two parallelograms are shown, with the smaller one having sides of 8 and 18, and the larger one having sides of 20 and x. The problem requires finding the length of side x by setting up a proportion: 8 / 18 = 20 / x. Solving this equation leads to x = 45. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms. Two parallelograms are shown, with the smaller one having sides of 8 and 18, and the larger one having sides of 20 and x. The problem requires finding the length of side x by setting up a proportion: 8 / 18 = 20 / x. Solving this equation leads to x = 45. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, with the smaller one having sides of 9 and 21, and the larger one having sides of x + 12 and 4x + 3. The problem requires setting up a proportion: 9 / 21 = (x + 12) / (4x + 3). Solving this equation leads to x = 15, which then allows us to find the side lengths of the larger parallelogram as 27 and 63. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, with the smaller one having sides of 9 and 21, and the larger one having sides of x + 12 and 4x + 3. The problem requires setting up a proportion: 9 / 21 = (x + 12) / (4x + 3). Solving this equation leads to x = 15, which then allows us to find the side lengths of the larger parallelogram as 27 and 63. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two triangles are similar using proportions. Two triangles are shown, both with a 70° angle. The first triangle has sides of 12 and 10, while the second has sides of 18 and 15. The problem requires setting up a proportion to check for similarity: 12 / 10 = 18 / 15. After simplifying, both ratios are equal (6 / 5 = 6 / 5), confirming that the triangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two triangles are similar using proportions. Two triangles are shown, both with a 70° angle. The first triangle has sides of 12 and 10, while the second has sides of 18 and 15. The problem requires setting up a proportion to check for similarity: 12 / 10 = 18 / 15. After simplifying, both ratios are equal (6 / 5 = 6 / 5), confirming that the triangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for b in a proportion where a = 8, c = 4, and d = 3. We set up the proportion 8 / b = 4 / 3 and solve for b, resulting in b = 6. This problem shows how to find an unknown value in the denominator of a proportion. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for b in a proportion where a = 8, c = 4, and d = 3. We set up the proportion 8 / b = 4 / 3 and solve for b, resulting in b = 6. This problem shows how to find an unknown value in the denominator of a proportion. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two triangles are not similar using proportions. Two triangles are shown, both with a 75° angle. The first triangle has sides of 15 and 9, while the second has sides of 28 and 18. The problem requires setting up a proportion to check for similarity: 15 / 9 = 28 / 18. After simplifying, the ratios are not equal (5 / 3 ≠ 14 / 9), concluding that the triangles are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two triangles are not similar using proportions. Two triangles are shown, both with a 75° angle. The first triangle has sides of 15 and 9, while the second has sides of 28 and 18. The problem requires setting up a proportion to check for similarity: 15 / 9 = 28 / 18. After simplifying, the ratios are not equal (5 / 3 ≠ 14 / 9), concluding that the triangles are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two right triangles are similar using proportions. Two right triangles are shown, one with legs of length 4 and 3, and the other with legs of length 10 and 7.5. The problem requires setting up a proportion to check for similarity: 4 / 3 = 10 / 7.5. After simplifying, both ratios are equal (4 / 3 = 4 / 3), confirming that the triangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two right triangles are similar using proportions. Two right triangles are shown, one with legs of length 4 and 3, and the other with legs of length 10 and 7.5. The problem requires setting up a proportion to check for similarity: 4 / 3 = 10 / 7.5. After simplifying, both ratios are equal (4 / 3 = 4 / 3), confirming that the triangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two right triangles are not similar using proportions. Two right triangles are shown, one with legs of length 12 and 5, and the other with legs of length 35 and 15. The problem requires setting up a proportion to check for similarity: 12 / 5 = 35 / 15. After simplifying, the ratios are not equal (12 / 5 ≠ 7 / 3), concluding that the triangles are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two right triangles are not similar using proportions. Two right triangles are shown, one with legs of length 12 and 5, and the other with legs of length 35 and 15. The problem requires setting up a proportion to check for similarity: 12 / 5 = 35 / 15. After simplifying, the ratios are not equal (12 / 5 ≠ 7 / 3), concluding that the triangles are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two rectangles are similar using proportions. Two rectangles are shown, with the first having dimensions 3 and 8, and the second having dimensions 9 and 24. The problem requires setting up a proportion to check for similarity: 3 / 8 = 9 / 24. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two rectangles are similar using proportions. Two rectangles are shown, with the first having dimensions 3 and 8, and the second having dimensions 9 and 24. The problem requires setting up a proportion to check for similarity: 3 / 8 = 9 / 24. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two rectangles are similar using proportions with algebraic expressions. Two rectangles are shown, with the first having dimensions 11.5 and 23, and the second having dimensions 23x and 46x. The problem requires setting up a proportion to check for similarity: 11.5 / 23 = 23x / 46x. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar for any value of x. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34TopicRatios, Proportions, and Percents DescriptionThis example illustrates how to determine if two rectangles are similar using proportions with algebraic expressions. Two rectangles are shown, with the first having dimensions 11.5 and 23, and the second having dimensions 23x and 46x. The problem requires setting up a proportion to check for similarity: 11.5 / 23 = 23x / 46x. After simplifying, the ratios are equal, confirming that the rectangles are indeed similar for any value of x. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are similar using proportions. Two parallelograms are shown, with the first having dimensions 6 and 9, and the second having dimensions 15 and 22.5. The problem requires setting up a proportion to check for similarity: 6 / 9 = 15 / 22.5. After simplifying, the ratios are equal, confirming that the parallelograms are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are similar using proportions. Two parallelograms are shown, with the first having dimensions 6 and 9, and the second having dimensions 15 and 22.5. The problem requires setting up a proportion to check for similarity: 6 / 9 = 15 / 22.5. After simplifying, the ratios are equal, confirming that the parallelograms are indeed similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are not similar using proportions. Two parallelograms are shown, with the first having dimensions 9 and 28, and the second having dimensions 18 and 54. The problem requires setting up a proportion to check for similarity: 9 / 18 ≠ 28 / 54. After simplifying, the ratios are not equal, concluding that the parallelograms are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36TopicRatios, Proportions, and Percents DescriptionThis example demonstrates how to determine if two parallelograms are not similar using proportions. Two parallelograms are shown, with the first having dimensions 9 and 28, and the second having dimensions 18 and 54. The problem requires setting up a proportion to check for similarity: 9 / 18 ≠ 28 / 54. After simplifying, the ratios are not equal, concluding that the parallelograms are not similar. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with side lengths 9 and 18, and the other with expressions 6x and 10x + 6. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 9 / 18 = 6x / (10x + 6). Solving this equation leads to x = 3. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar triangles with algebraic expressions. Two triangles are shown, one with side lengths 9 and 18, and the other with expressions 6x and 10x + 6. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 9 / 18 = 6x / (10x + 6). Solving this equation leads to x = 3. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with side lengths 4 and 3, and the other with expressions 3x and 2x + 1. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 4 / 3 = 3x / (2x + 1). Solving this equation leads to x = 4. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar right triangles with algebraic expressions. Two right triangles are shown, one with side lengths 4 and 3, and the other with expressions 3x and 2x + 1. The problem requires setting up a proportion to determine the value of x for which the triangles are similar: 4 / 3 = 3x / (2x + 1). Solving this equation leads to x = 4. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with an unknown side length. Two rectangles are shown, one with side lengths of 4 and x, and the other with side lengths of 15 and 4. The problem requires setting up a proportion to determine the value of x for which the rectangles are similar: 4 / 15 = x / 4. Solving this equation leads to x = 16/15. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving a proportion problem using similar rectangles with an unknown side length. Two rectangles are shown, one with side lengths of 4 and x, and the other with side lengths of 15 and 4. The problem requires setting up a proportion to determine the value of x for which the rectangles are similar: 4 / 15 = x / 4. Solving this equation leads to x = 16/15. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for b in a proportion where a is expressed as x + 2, and c and d are given constants (c = 5, d = 2). We set up the equation (x + 2) / b = 5 / 2 and solve for b, resulting in the expression b = (2(x + 2)) / 5. This problem demonstrates how to handle algebraic expressions in proportions. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for b in a proportion where a is expressed as x + 2, and c and d are given constants (c = 5, d = 2). We set up the equation (x + 2) / b = 5 / 2 and solve for b, resulting in the expression b = (2(x + 2)) / 5. This problem demonstrates how to handle algebraic expressions in proportions. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving a proportion problem using similar parallelograms with algebraic expressions. Two parallelograms are shown, one with side lengths of 6 and x, and the other with expressions of 3x + 2 and 22. The problem requires setting up a proportion to determine the value of x for which the parallelograms are similar: 6 / 22 = x / (3x + 2). Solving this equation leads to x = 3. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5TopicRatios, Proportions, and Percents DescriptionThis example demonstrates solving for c in a proportion where a = 9, b = 4, and d = 12. We set up the proportion 9 / 4 = c / 12 and solve for c, resulting in c = 27. This problem shows how to find an unknown value in the numerator of a proportion. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion. |
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Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 | Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6TopicRatios, Proportions, and Percents DescriptionThis example illustrates solving for c in a proportion where a = 8, b = 3, and d is expressed as x + 3. We set up the equation 8 / 3 = c / (x + 3) and solve for c, resulting in the expression c = (8(x + 3)) / 3. This problem demonstrates how to handle algebraic expressions in the denominator of a proportion. |
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