Change of Subject In Mathematics

“Change of subject” is a term used in algebra when we want to rearrange an equation so that a different variable (or subject) is isolated on one side of the equation. This is a useful technique that allows us to solve equations and find the value of a particular variable.

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Let me give you five examples to help illustrate this concept:

Example 1: Suppose we have the equation: 2x + 3 = 9. We want to solve for x, which means we need to isolate x on one side of the equation. To do this, we can use the “change of subject” technique to rearrange the equation: 2x = 9 – 3 2x = 6 x = 3

Example 2: Suppose we have the equation: 5y – 7 = 13. We want to solve for y. To do this, we can use the “change of subject” technique to rearrange the equation: 5y = 13 + 7 5y = 20 y = 4

Example 3: Suppose we have the equation: 4z – 6 = 10z. We want to solve for z. To do this, we can use the “change of subject” technique to rearrange the equation: 4z = 10z + 6 4z – 10z = 6 -6z = 6 z = -1

Example 4: Suppose we have the equation: 2a + 3b = 12. We want to solve for a. To do this, we can use the “change of subject” technique to rearrange the equation: 2a = 12 – 3b a = (12 – 3b)/2

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Example 5: Suppose we have the equation: 2x + 3y – 5 = 0. We want to solve for y. To do this, we can use the “change of subject” technique to rearrange the equation: 3y = -2x + 5 y = (-2x + 5)/3

Evaluation

1. Which of the following is the best way to “change the subject” in an algebraic equation? a) Add the same value to both sides of the equation b) Subtract the same value from both sides of the equation c) Multiply both sides of the equation by the same value d) Divide both sides of the equation by the same value
2. Suppose we have the equation 3x + 4 = 10. What is the value of x when we “change the subject” and solve for x? a) 2 b) 3 c) 4 d) 5
3. Suppose we have the equation 5y – 8 = 17. What is the value of y when we “change the subject” and solve for y? a) 3 b) 5 c) 6 d) 9
4. Suppose we have the equation 4x – 6y = 18. What is the value of y when we “change the subject” and solve for y? a) y = (4x – 18)/6 b) y = (18 – 4x)/6 c) y = (4x + 18)/6 d) y = (18 + 4x)/6
5. Suppose we have the equation 2a + 3b = 12. What is the value of a when we “change the subject” and solve for a? a) a = (12 – 3b)/2 b) a = (3b – 12)/2 c) a = (12 + 3b)/2 d) a = (3b + 12)/2
6. Suppose we have the equation 6x + 9 = 21. What is the value of x when we “change the subject” and solve for x? a) 1 b) 2 c) 3 d) 4
7. Suppose we have the equation 4y – 7 = 9. What is the value of y when we “change the subject” and solve for y? a) 4 b) 3 c) 2 d) 1
8. Suppose we have the equation 2x – 3 = 7x. What is the value of x when we “change the subject” and solve for x? a) x = -1/5 b) x = -1/2 c) x = -2/5 d) x = -3/2
9. Suppose we have the equation 5a – 6b = 9. What is the value of b when we “change the subject” and solve for b? a) b = (5a – 9)/6 b) b = (9 – 5a)/6 c) b = (5a + 9)/6 d) b = (9 + 5a)/6
10. Suppose we have the equation 3x + 2y = 16. What is the value of y when we “change the subject” and solve for y? a) y = (16 – 3x)/2 b) y = (3x – 16)/2 c) y = (16 + 3x)/2 d) y = (3x + 16)/2

Lesson Objective: By the end of the lesson, students will be able to “change the subject” in an algebraic equation to isolate a variable.

Materials Needed:

• Whiteboard and markers
• Algebraic equations on handouts or slides
• Worksheet for students to practice “changing the subject”

Lesson Outline:

1. Introduction (5 minutes)

• Greet students and introduce the topic of “change of subject” in algebra
• Explain that this is a technique used to rearrange an equation to isolate a specific variable
• Give an example of why this is useful (for example, to solve for an unknown value in an equation) [mediator_tech]

2. Demonstration (10 minutes)

• Write an equation on the board (for example, 2x + 5 = 15)
• Show students how to “change the subject” and isolate x on one side of the equation (in this case, x = 5)
• Repeat with a few more equations, allowing students to ask questions and follow along

3. Guided Practice (10 minutes)

• Hand out a worksheet with several equations on it
• Ask students to work in pairs to “change the subject” and solve for the variable in each equation
• Walk around the room to answer questions and provide assistance as needed

4. Independent Practice (15 minutes)

• Give students another worksheet with more complex equations to solve
• Ask students to work independently to solve the equations and “change the subject”
• Collect the worksheets for assessment

5. Conclusion (5 minutes)

• Review the steps for “changing the subject” in an algebraic equation
• Remind students of the importance of this technique in solving equations and finding unknown values
• Answer any remaining questions and dismiss the class

Assessment: Assessment will be based on the completion of the independent practice worksheet, as well as observation during the guided practice portion of the lesson.

Differentiation: For students who are struggling with the concept, provide additional guided practice and examples. For students who are ready for more advanced work, provide more complex equations to solve or challenge them to create their own equations to solve

Weekly Assessment /Test

1. To “change the subject” in an equation, we want to isolate a specific _________ on one side of the equation.
2. When we “change the subject,” we can add or subtract the same value to ________ both sides of the equation.
3. If we want to “change the subject” and solve for y in the equation 4x + 3y = 12, we need to move the 4x to the other side of the equation by __________ it from both sides.
4. When we “change the subject,” we can also multiply or divide both sides of the equation by the same __________.
5. If we want to “change the subject” and solve for x in the equation 2x – 3y = 7, we need to move the -3y to the other side of the equation by __________ it from both sides.
6. If we want to “change the subject” and solve for z in the equation 2z/3 + 4 = 8, we need to first __________ both sides of the equation by 3/2.
7. To “change the subject” and solve for b in the equation 5a – 6b = 9, we need to divide both sides of the equation by __________. [mediator_tech]
8. When “changing the subject,” it’s important to perform the same operation on both sides of the equation to ensure that the equation remains __________.
9. If we want to “change the subject” and solve for y in the equation 3x + 2y = 16, we need to move the 3x to the other side of the equation by __________ it from both sides.
10. To “change the subject” and solve for a in the equation 2a + 3b = 12, we need to subtract 3b from both sides of the equation, and then __________ both sides by 2.

1. Solve for y in 3x + 2y = 15:
   a) y = (15/2) – 3x
   b) y = (15 – 3x)/2
   c) y = (3x – 15)/2
   d) y = 5 – 2x

2. Find z in 4a – 2b + z = 10:
   a) z = 10 – 4a + 2b
   b) z = 4a – 2b – 10
   c) z = 10 – 2b – 4a
   d) z = 4a + 2b + 10

3. Solve for b in 2ab + c = 5b:
   a) b = (5 – c)/(2a)
   b) b = (2a – c)/5
   c) b = (5/(2a)) – c
   d) b = (5c – 2a)/c

4. Find r in A = 3πr^2:
   a) r = sqrt(A/(3π))
   b) r = A/(3π)
   c) r = sqrt(A)/3π
   d) r = (3π)/A

5. Solve for x in 5y – 2x = 8:
   a) x = (5y – 8)/2
   b) x = (8 – 5y)/2
   c) x = 4 – (5y/2)
   d) x = (8 + 5y)/2

6. Find a in 2ab – c = 3a:
   a) a = c/(2b – 3)
   b) a = (2b – 3)/c
   c) a = (2b – 3)/3
   d) a = 3/(2b – c)

7. Solve for y in 4x + 2y = 10:
   a) y = (10 – 4x)/2
   b) y = 5 – 2x
   c) y = 2 – (4x/5)
   d) y = 10/2 – x

8. Find z in 3a – 2b + z = 15:
   a) z = 15 – 3a + 2b
   b) z = 2b – 3a + 15
   c) z = 3a – 15 – 2b
   d) z = 15 + 2b – 3a

9. Solve for b in 2ab + 4 = 8b:
   a) b = (2a + 4)/8
   b) b = (4 – 2a)/8
   c) b = (2a – 4)/8
   d) b = 8/(2a + 4)

10. Find r in A = 2πr^2:
    a) r = sqrt(A/(2π))
    b) r = A/(2π)
    c) r = sqrt(A)/2π
    d) r = (2π)/A

11. Solve for x in 3y – 2x = 12:
    a) x = (3y – 12)/2
    b) x = 6 – (3y/2)
    c) x = (12 – 3y)/2
    d) x = 12/2 – (3y/2)

12. Find a in 3ab – c = 2a:
    a) a = c/(3b – 2)
    b) a = (3b – 2)/c
    c) a = (3b – 2)/2
    d) a = 2/(3b – c)

13. Solve for y in 4x + 2y = 16:
    a) y = (16 – 4x)/2
    b) y = 8 – 2x
    c) y = 2 – (4x/8)
    d) y = 16/2 – 2x

14. Find z in 2a – 3b + z = 7:
    a) z = 7 – 2a + 3b
    b) z = 3b – 2a + 7
    c) z = 2a – 7 – 3b
    d) z = 7 + 3b – 2a

15. Solve for b in 3ab + 5 = 7b:
    a) b = (3a + 5)/7
    b) b = (5 – 3a)/7
    c) b = (3a – 5)/7
    d) b = 7/(3a + 5)