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Illustrative Math-Media4Math Alignment

 

 

Illustrative Math Alignment: Grade 7 Unit 7

Expressions, Equations, and Inequalities

Lesson 12: Solving Problems about Percent Increase or Decrease

Use the following Media4Math resources with this Illustrative Math lesson.

Thumbnail Image Title Body Curriculum Topic
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 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 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 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
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 2 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 2 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 2

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving a proportion where b is expressed as x + 1, and c and d are given constants (c = 3, d = 2). The goal is to solve for a using the proportion a / b = c / d. By substituting the known values, we set up the equation a / (x + 1) = 3 / 2 and solve for a, resulting in the expression a = (3(x + 1)) / 2.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 20

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 21

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 22

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 23

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 24

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 25

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 26

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 27

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 28

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 29

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 3

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 30

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 31

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 32

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 33

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 34

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 35

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 36

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 37

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 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 38

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 39

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 4

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 40

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 5

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 6

Topic

Ratios, Proportions, and Percents

Description

This 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.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 7

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving for d in a proportion where a = 12, b = 5, and c = 48. We set up the proportion 12 / 5 = 48 / d and solve for d, resulting in d = 20. This problem shows how to find an unknown value in the denominator of a proportion when all other values are known.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 8

Topic

Ratios, Proportions, and Percents

Description

This example illustrates solving for d in a proportion where a = 11, b = 5, and c is expressed as x - 4. We set up the equation 11 / 5 = (x - 4) / d and solve for d, resulting in the expression d = (5(x - 4)) / 11. This problem demonstrates how to handle algebraic expressions in the numerator of a proportion.

Proportions
Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9 Math Example--Ratios, Proportions, and Percents--Solving Proportions: Example 9

Topic

Ratios, Proportions, and Percents

Description

This example demonstrates solving a proportion problem using similar triangles. Two triangles are shown, with one having sides of 3 and 4, and the other having sides of 6 and x. The problem requires finding the length of side x by setting up a proportion based on the similar triangles: 3 / 4 = 6 / x. Solving this equation leads to x = 8.

Proportions
Math Example--Solving Equations--One-Variable Equations: Example 1 Math Example--Solving Equations--One-Variable Equations: Example 1 One-Variable Equations: Example 1

Topic

Equations

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 10 Math Example--Solving Equations--One-Variable Equations: Example 10 One-Variable Equations: Example 10

Topic

Equations

Description

This example involves solving a one-variable equation that may include complex terms or require multiple steps to simplify. The equation might involve fractions, decimals, or variables on both sides. Solving it involves using inverse operations, distributing terms, and combining like terms to isolate the variable. This type of problem helps students refine their algebraic skills and understand the importance of systematic problem-solving. Checking the solution by substituting it back into the original equation is a crucial step to ensure accuracy.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 11 Math Example--Solving Equations--One-Variable Equations: Example 11 One-Variable Equations: Example 11

Topic

Equations

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 12 Math Example--Solving Equations--One-Variable Equations: Example 12 One-Variable Equations: Example 12

Topic

Equations

Description

This example presents a one-variable equation that may involve variables on both sides. The solving process requires moving all terms involving the variable to one side and constants to the other. This often involves using the distributive property and combining like terms. The goal is to isolate the variable and solve for its value. This type of problem helps students develop their algebraic manipulation skills and understand the importance of maintaining balance in an equation. Checking the solution by substituting it back into the original equation is crucial to ensure accuracy.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 13 Math Example--Solving Equations--One-Variable Equations: Example 13 One-Variable Equations: Example 13

Topic

Equations

Description

This example involves solving a one-variable equation that might include complex expressions, such as those with parentheses or multiple terms. The solving process may require using the distributive property to eliminate parentheses and combining like terms to simplify the equation. After simplification, standard techniques are used to isolate the variable. This example reinforces the importance of following the order of operations and checking the solution for accuracy. Mastery of these skills is essential for tackling more advanced algebraic problems.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 14 Math Example--Solving Equations--One-Variable Equations: Example 14 One-Variable Equations: Example 14

Topic

Equations

Description

This example deals with solving a one-variable equation that may involve more complex algebraic expressions, such as nested parentheses or fractional coefficients. The solving process requires careful application of the distributive property and combining like terms. After simplifying the equation, inverse operations are used to isolate the variable. This example highlights the importance of precision in algebraic manipulation and the necessity of verifying solutions by substituting them back into the original equation to ensure correctness.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 15 Math Example--Solving Equations--One-Variable Equations: Example 15 One-Variable Equations: Example 15

Topic

Equations

Description

This example involves solving a one-variable equation that may include complex terms or require multiple steps to simplify. The equation might involve fractions, decimals, or variables on both sides. Solving it involves using inverse operations, distributing terms, and combining like terms to isolate the variable. This type of problem helps students refine their algebraic skills and understand the importance of systematic problem-solving. Checking the solution by substituting it back into the original equation is a crucial step to ensure accuracy.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 16 Math Example--Solving Equations--One-Variable Equations: Example 16 One-Variable Equations: Example 16

Topic

Equations

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 17 Math Example--Solving Equations--One-Variable Equations: Example 17 One-Variable Equations: Example 17

Topic

Equations

Description

This example presents a one-variable equation that may involve variables on both sides. The solving process requires moving all terms involving the variable to one side and constants to the other. This often involves using the distributive property and combining like terms. The goal is to isolate the variable and solve for its value. This type of problem helps students develop their algebraic manipulation skills and understand the importance of maintaining balance in an equation. Checking the solution by substituting it back into the original equation is crucial to ensure accuracy.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 18 Math Example--Solving Equations--One-Variable Equations: Example 18 One-Variable Equations: Example 18

Topic

Equations

Description

This example involves solving a one-variable equation that might include complex expressions, such as those with parentheses or multiple terms. The solving process may require using the distributive property to eliminate parentheses and combining like terms to simplify the equation. After simplification, standard techniques are used to isolate the variable. This example reinforces the importance of following the order of operations and checking the solution for accuracy. Mastery of these skills is essential for tackling more advanced algebraic problems.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 19 Math Example--Solving Equations--One-Variable Equations: Example 19 One-Variable Equations: Example 19

Topic

Equations

Description

This example deals with solving a one-variable equation that may involve more complex algebraic expressions, such as nested parentheses or fractional coefficients. The solving process requires careful application of the distributive property and combining like terms. After simplifying the equation, inverse operations are used to isolate the variable. This example highlights the importance of precision in algebraic manipulation and the necessity of verifying solutions by substituting them back into the original equation to ensure correctness.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 2 Math Example--Solving Equations--One-Variable Equations: Example 2 One-Variable Equations: Example 2

Topic

Equations

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations
Math Example--Solving Equations--One-Variable Equations: Example 20 Math Example--Solving Equations--One-Variable Equations: Example 20 One-Variable Equations: Example 20

Topic

Equations

Description

This example involves solving a one-variable equation that may include complex terms or require multiple steps to simplify. The equation might involve fractions, decimals, or variables on both sides. Solving it involves using inverse operations, distributing terms, and combining like terms to isolate the variable. This type of problem helps students refine their algebraic skills and understand the importance of systematic problem-solving. Checking the solution by substituting it back into the original equation is a crucial step to ensure accuracy.

Solving Multistep Equations, Solving One-Step Equations and Solving Two-Step Equations