## Thursday, 20 February 2014

### 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.

1. (3a+b) is not (3ab) because (3ab) is (3 x a x b) not (3a + b).
(2s + 4t) is not (6st) because (6st) is (6 x s x t)

2. What's the thinking in Jane's head?

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4. jane was thinking of multiplying the numbers

5. Jane was thinking of combining them together variable a and b together. But she forgot that 3ab is 3a and 3b. In this case, b is b, not 3b. For the second equation, she also made the same mistake. The number used to times the variable is different so she cannot combine them therefore the workings are incorrect.

6. Jane was thinking that 2+4=6 and s+t=st. This only applies when you multiply the numbers

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7. Jane is wrong.She thinks that 3ab= 3a +b but 3ab= 3a x b so Jane is wrong.
She thinks that 6st=2s +4t but 6st= 2s x 4t.

8. Jane was thinking about multiplying 3a and b when she did 3a + b.

9. This is not true, because when you cannot add two different "letters". You can only multiply different "letters", in this case 3a+b = 3ab would be wrong, 3a x b = 3ab. She was probably thinking that the letters can be add or subtracted.

10. For, 3a + b = 3ab, 3ab is 3 x a x b, but she has instead added the b to 3 x a.

11. 3a+b is not equal to 3ab because 3ab is 3 x a x b.
2s+4t is not equal to 6st because 6st is 6 x s x t

She must be thinking 3a + b is 3ab because she might be thinking 3a plus b, just add b to 3a(b)
She must be thinking 2s + 4t is 6 st as she might be thinking 2+4 =6, and s+t =st

12. Jane was thinking of combining the two variables 'a' and 'b' but then she forgot that 'a' and 'b' are different variables just like for the second equation

13. No, Jane is wrong.

3a+b = (3 x a) + (1 x b)
while 3ab means 3 x a xb

2s+4t= (2 x s) + (4 x t)
while 6st means 6 x s x t

In Jane's head, she was probably thinking that 3a+b means multiplying b to 3a, and 2s+4t means adding adding (2+4) x (s x t)

14. This is not true. Jane probably thought that she could add 2 different letters when you actually cannot.

15. She was thinking that she has to do what she does in multiplication for part b where she is doing 2+4 and s+t

16. Jane mistook addition of 3a and b with multiplying 3a and b. If the addition symbol was replaces with a multiplication symbol, her working would be correct.

17. (3a+b) is not (3ab) because (3ab)is equal to (3 multiplied by a multiplied by b).
(2s+4t)is not (6st) because (6st)is equal to (6 multiplied by s multiplied by t)
Jane might be thinking that it is appropriate to add the different expressions together like it is done in multiplication in algebra.

18. *the different variables*

19. It can be both wrong and correct as if a is 1 and b is 1.5 then question one is possible.

20. 3ab is=3x(axb)
and not
3a+b

21. both can be correct as for the 1st one, a can be b/3(b-1) and b can be 3a/3a-1. for the 2nd one, s can be 2t/3t-1 and t can be s/3s-2

22. 3a + b = 3ab is wrong.
Jane thought that a number beside a letter means addition, but it actually multiplication.

2s + 4t = 6st
Jane thought that she could add the "2" and "4" together as a non-algebraic number. She does not know that the algebraic numbers are different units.

23. She thinks that 3a+b is 3ab but it is 3axb.
She thinks that 2s+4t is 6st but it is 6sx4t.

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