Chap 4: Can you tell where has gone wrong?

Textbook (p89)

Julie says that
3a² + 2a + 5a² = 10a³
and
7b x 5b² x 6 = 210b
Do you agree with Julie? If not, what are the mistakes in her algebraic manipulations?

Enter your responses under Comments.

Note: to enter superscripts, you may use the code

  • & sup 2 ; (type them without space) to raise to the power of 2
  • & sup 3 ; (type them without space) to raise to the power of 3

15 comments:

  1. I do not agree with Julie.

    It should be :
    3a² + 2a + 5a²
    = (3a² + 5a²) + 2a
    = 8a² + 2a(Ans.)
    3a² and 5a² cannot be added with 2a as they are of different "power".

    It should be:
    7b x 5b² x 6
    = (7b x 5b²) x 6
    = 35b³ x 6
    = 210b³(Ans.)
    b x b² = b³
    Julie probably took b² ÷ b = b. Therefore, her final answer is 210b without the "power" of 3.

    JJ(:

    ReplyDelete
  2. I don't agree with her.
    1) 3a² and 5a² cannot be added together with 2a. That is because they have different "power", like 2a and 2a².
    It should be: 3a² + 2a + 5a²
    = (3a² + 5a²) + 2a
    = 8a² + 2a

    2) Julie took the amount with the "power" like a fraction, so it became b² ÷ b, which lead her to the answer, 210b, without the "power of 3" behind.
    It should be: 7b x 5b²
    = 35b³
    35b³ x 6
    = 210b³

    ReplyDelete
  3. This comment has been removed by the author.

    ReplyDelete
  4. 1) 3a&sup2 + 2a + 5a&sup2
    =(3 x a x a) + (2 x a) + (5 x a x a)
    =(8 x a x a) + (2 x a)
    =8a&sup2 + 2a

    2) 7b x 5b&sup2 x 6
    =(7 x b) x (5 x b x b) x 6
    =(7 x 5) x (b x b x b) x 6
    =35b&sup3 x 6
    =210b&sup3

    Both numbers have different powers. There are like 8 groups of a&sup2 and 2 groups of a. But the a and a&sup2 are different powers so they cannot be grouped together. Therefore Julie's first answer is wrong.

    As for her second answer, which is also incorrect, she probably "grouped" the "b"s together and treated them as one variable, so her answer was not to the power of 3.

    ReplyDelete
  5. This comment has been removed by the author.

    ReplyDelete
  6. I do not agree with Julie.

    For number one, the workings are suppose to be

    = 3a² + 2a + 5a²
    = (3a² + 5a²) + 2a <------- Only equivalent units can be added together.
    = (8a² + 2a)<-------------Answer

    So basically, julie just added all the "powers" and the digits together to form a irregular expression.

    For question two,

    The answer should be

    =7b x 5b² x 6
    = (7b x 5b²)x 6<------Bracketed to make the expression clearer
    = 35b³ x 6<-------- Note that b x b&sup2 = b&sup3
    = 210b³<----------Answer


    Basically what julie did is that she made all the "b" into one variable so the b&sup3 will be grouped into a single "b" thus leading to her mistakes in her calculations

    ReplyDelete
  7. I do not agree with Julie.

    For the first question, it should be 3a&sup2+2a+5a&sup2=8a&sup2+2a.
    Julie added the numbers, 3,2 and 5, together and supposed that a&sup2+a=a&sup3.

    For the second question, it should be 7bx5b&sup2x6=210b&sup3. Julie only multiplied the numbers and only used b without using b&sup2. So she supposed that 7bx5b&sup2x6=210b.

    ReplyDelete
  8. I do not agree with Julie.

    Question 1 should be:
    3a²+2a+5a²=(3a²+5a²)+2a
    =8a²+2a
    Julie most probably added 2a to 8a². They cannot be added together as they are of different values.

    Question 2 should be:
    7bx5b²x6=35b³x6
    =210b³
    Julie most likely thought that b² was the same as b and grouped them under the same value.

    ReplyDelete
  9. I do not agree with Julie

    1a) 3a² + 2a + 5a² = 10a³

    3a² +2a + 5a²
    =(3a² +5a² )+ 2a
    =(8a² + 2a)

    2a)7b x 5b² x 6 = 210b
    =35b7² x6
    =210³

    ReplyDelete
  10. This comment has been removed by the author.

    ReplyDelete
  11. This comment has been removed by the author.

    ReplyDelete
  12. I do not agree with Julie.

    For question 1, 3a^2+2a+5a^2 cannot be added as they are of different values. It should be (3a^2+5a^2)+2a, which will give us 8a^2+2a.

    For question 2, 7bx5b^2x6 should equal to 35b^3x6, which gives us 210b^3

    ReplyDelete
  13. I do not agree with Julie.

    Question 1:
    3a²+ 2a+ 5a²=(3a²+5a²) +2a
    = 8a²+ 2a
    Julie might have added 2a to 8a². They can't be added together as they are of different values.

    Question 2:
    7bx5b²x 6=35b³x6
    = 210b³
    Julie might have thought that b² was the same as b and grouped them under the same value.

    ReplyDelete
  14. I do not agree Julie.

    Question 1


    3a² + 2a + 5a² = 3a x a + 2a + 5a x a
    = 8a²+2a



    7b x 5b² x 6 = 7b x 5b x b x 6
    = (7 x 5 x 6 )( b x b x b)
    = 210b³

    ReplyDelete
  15. I do not agree with Julie.

    It should be:

    3a² + 2a + 5a² = 8a² + 2a

    7b x 5b² x 6 = 210b³

    ReplyDelete