Chap 4 Algebra... Help! What's Wrong?

Jane did the following:
  • 3a + b = 3ab
  • 2s + 4t = 6st
Do you think Jane is correct in her algebraic manipulations?
If yes, please write down examples to show that her answer is correct.

If not, explain to Jane her mistakes and help her to correct.

Enter your response in Comments (this is part of your daily work...)

18 comments:

  1. No, Jane is not correct in her algebraic manipulations.

    The sign is addition sign but when Jane puts a and b together as "ab" it means a X b which will not be the same value anymore.

    It applies the same to the second statement.

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  2. Jane is not correct.
    Putting a and b together indicates that a is multiplied by b.
    Putting s and t together indicates that s is multiplied by t

    Since this is addition, the statements should be:
    3a + b
    and
    2s + 4t

    This is because they cannot be simplified further.

    ReplyDelete
  3. Jane is not correct in her calculations


    Placing "a" and "b" together is equivalent to multiplying the two unknowns.

    In the same context, placing "s" and "t" together is equivalent to multiplying the two unknowns.


    If she wants to place it in an addition format, Jane should write the expression like this

    3a+b=3a+b

    and

    2s+4t=2s+4t

    ReplyDelete
  4. Jane is not correct in her calculations.

    Placing A and B together is saying that she is expressing that a and b is multiplied. This applies to the 2nd statement.

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  5. Jane is not correct.

    3a+b=(3a+b)
    Not 3ab

    2s+4t=(2s+4t)
    Not 6st

    The a and b are different constants just like s and t

    ReplyDelete
  6. No, Jane is not correct.

    Doing this, she is interpreting that "a" is multiplied by "b", and that "s" is multiplied by "t", which is not correct.

    For the 1st statement, "3a" and "b" are two different things. Therefore, they cannot be added together. It should be left as "3a+b".

    For the 2nd statement, "2s" and "4t" are two different things. Therefore they cannot be added together. It should be left as "2s+4t"

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  7. Jane is not correct in her calculations. In example 1, a and b are two different figures, therefore they cannot be added together as though they are of the same figure. The same thing applies for the second sum.

    The correct answer should be (3a + b) and (2s + 4t).

    ReplyDelete
  8. Jane is not correct in her algebraic manipulations.

    3a + b = 3a + b or b + 3a
    It is not suppose to be 3ab. 3ab= 3a x b, which is wrong.

    2s + 4t = 2s + 4t or 4t + 2s
    It is not suppose to be 6ts. 6ts= 4t x 2s, which is wrong.

    JJ(:

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  9. 3a + b is not 3ab.
    3ab is = 3a x b
    2s + 4t is not 6ts.
    6ts is = 2s x 4t

    ReplyDelete
  10. No. different algebraic expressions cannot be added together. thus the answer should be:
    3a + b = 3a+b
    2s + 4t =2s+4t

    ReplyDelete
  11. To:
    Ziying
    Joshua (MasterChief)
    Calvin
    Cheng Ngee
    Jonathan
    Shamus
    Jianhui
    Jing Jie
    Zhiqi
    Grace

    To prove something that is incorrect, the easiest way is to use some numbers to show the mistake.
    Hint: Let "a" and "b" be 2 numbers like "1" and "7", find out what 3a+b gives, and what 3ab gives... then do a comparison :D

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  12. She is wrong.
    Let a be 3, b be 5, s be 7 and t be 9.

    So, if we calculate, it should be: 3a + b = (3x3)+5
    = 14
    And : 2s + 4t = 2x7 + 4x9
    = 14 + 36
    =50
    Not 3ab which is actually 3a x b
    and not 6st which is 2+4 x (sxt)

    ReplyDelete
  13. No, Jane us not correct as only 3a x b would give you 3ab . If it is 3a + b, the answer should be the same as the statement.
    This concept applies to the next algebraic manipulations.

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  14. No, Jane is incorrect. To get both answers, she would have to multiply the two numbers in the statements instead of adding them together.

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  15. No, Jane is wrong. Both numbers are represented by different variables. When they are represented by different variables, they are of different value and therefore cannot be added together.

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  16. Guang Jun, Soh Fan and Joshua,
    In a way, you have rightly pointed out the error in the operations.

    Now, let's consider Zhi Chao's response.
    He mentioned, "both numbers are represented by different variables".

    This leads to another 'discussion':
    What if...
    variables a and b represent the same number. Will the first equation become true?

    What if...
    variables s and t represent the same number,
    Will the second equation become true?

    ReplyDelete
  17. Jane did the following:
    3a + b = 3ab
    2s + 4t = 6st
    Do you think Jane is correct in her algebraic manipulations?
    If yes, please write down examples to show that her answer is correct.

    If not, explain to Jane her mistakes and help her to correct.

    No, Jane is not correct in her algebraic manipulation.

    If Jane says 3a + b = 3ab, then she meant the answer is equal to 3 x a x b. But, that isn't what the question is looking for. So, the correct answer should be 3a + b = 3a + b.

    If Jane says 2s + 4t = 6st, then she meant the answer is equal to 6 x s t. But, that isn't what the question is looking for. So, the correct answer should be 2s + 4t = 2s + 4t.

    ReplyDelete
  18. Even if variables a and b represent the same number, the first equation is still wrong.
    Let a and b be 5,
    (3x5)+5=20
    3a + a (since a and b are the same)=4a
    4a=4x5=20

    Even if variables s and t represent the same number, the second equation is still wrong.
    Let s and t be 3,
    (2x3)+ (4x3) =18
    2s + 4s = 6s
    6s= 6x3 = 18

    So, Jane is wrong.

    ReplyDelete