## Monday, April 18, 2011

### Chapter 4: Algebra... Help! What's Wrong?

Given a and b are 2 unlike terms.
It is given that

• 3a + b
• 2s + 4t

Jane did the following algebraic manipulation:
• 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.
[you may substitute values for both a and  to prove your case]

1. No.
3a+b = (3a)+b
This means its 3 multiplied by a before adding b.
Her answer, 3ab, can be shown as 3 x a x b or 3 times a times b which is far from (3a)+b.
Same applies for 2s+4t just that the values are different.

2. No.
to get 3ab the equation should be 3axb.
The question does not show a times sign,instead a plus. This also applies to 2s+4t.

3. No.
The type of things are represented by different alphabets, which are not the same. It is the same as trying to add 2 apples and 4 oranges which are not the same thing.

4. No . For 3a + b = 3ab , it should be (3 x a) + b , not 2ab . For 2s + 4t = 6st , the same thing applies . (2 x s) + (4 x t)

5. Nope, SHE'S WRONG. 3ab is equal to 3xaxb. But the question is asking for (3xa)+b. She should put her answer as: 3a+b
2s+4t does not equal to 6st. 2s+4t = (2xs)+(4xt)
6st= (6 x s x t). So 2s+4t does not equal to 6st

6. No,as 3a+b=3a+b,as the algebraic expressions and value are different and cannot be added to each other in one simple expression

7. no. if she says 3ab, it is actually 3xaxb bit the question is 3xa + b. the same concept is applied to the next question

8. No, Jane is wrong in her algebraic manipulations. If the sign says add you cannot change it to a multiplication.
For example:
If a=5 b=10
3(5)+10= 25
However in Jane's formula.
(3)(5)(10)=150
The answers are different, so it is wrong.

9. No.
In the 1st situation, Jane was suppose to find out 3a + b. The answer deduced was 3ab. That is incorrect as 3ab is suppose to be 3 x a x b. Which totally differs from the correct answer (3a +b).
2nd Situation : She was asked to find 2s + 4t. The answer she gave was 6st. 6st is incorrect as it is 6 x s x t. Which it totally different from the correct answer which should be (2s + 4t).

In both situations, as the unknown numbers (a & b) and (s & t) must be unlike numbers, it is not possible to find and answer that is alike for both the correct and incorrect answers in both scenario.

10. No. She is saying that b is multiplied of 3 instead of adding with 3 + a.This also applies 2s+4t. It is two different values are different
Example:
if..
a=5
b=2
3(5)+2=17

accordinng to jane's method=3(5)(3)=24
this proves that jane's answer is wrong.

11. No.
3a=a+a+a
So if a+a+a+b, it will still remain as 3a+b
2s=s+s;4t=t+t+t+t
So if s+s+t+t+t+t, it will also remain as 2s+4t

12. No,
a) is (3a+b)
b) is (2s+4t)

13. No, Jane is wrong.
3a is a+a+a. So, 3a+b is (a+a+a)+b. Since (a) and (b) are different values, they cannot be added together. As a result, 3a+b will still be written in the same manner.
The same concept is applied to 2s+4t.
2s is s+s. 4t is t+t+t+t. 2s+4t is then (s+s)+(t+t+t+t). (s) and (t) are different values and cannot be added together. As a result, 2s+4t is still written as 2s+4t.

14. a)3a is three times of a, it is multiplied before the b is added.
b) the s and the t cannot be combined as one like that. They are separate, so they have to be counted separately. The 2s are one group while the 4t is another

15. No, she is wrong.

a )

3a + b = a + a + a + b

a and b are unlike items, so it will remain as 3a + b

b )

2s + 4t = s + s + t + t + t + t

s and t are unlike items, so it will remain as 2s + 4t

16. No. Both statements are wrong.
Since a and b are different, they shouldn't be added together like that. It will change the whole value. Her answer should be left alone as (3a+b).
The same applies to the second statement as well, just that the answer should be (2s+4t) instead.

17. No, Jane is wrong.
3a + b = 3ab WRONG
(3a) + b = (3 x a) + b = (3a+b) CORRECT
a and b are of different values. So, 3a+b cannot be put in 3ab because it does not show 3a + b, it shows 3 x a x b.
2s + 4t = 6st WRONG
(3a) + (4t) = (2 x s) + (4 x t) = (2s+4t) CORRECT
s and t are of different values. So, 2s+4t cannot be put in 6st because it does not show 2s + 4t, it shows 6 x s x t.

18. No , she is wrong in both statements.
3a+b is not equal to 3ab
It is supposed to be (3a+b)
And the same thing applies to the other one.