### Chapter 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.

1. No Jane is wrong. As for example, 4a, it means 4(a) and 4a(b) = to 4ab, but 4+a= (4+a) as the value of a is unknown. hence 4+a+b=(4+a+b)

2. No.

1. (ab) actually means that a*b. In this case, it is (a+b), thus her expressions are wrong. It should have been: 3a+b= (3a+b)

2. In this case, (s) cannot be multiplied with (t) to get (st). This expression is about adding the numbers up. Furthermore, (2s) and (4t) are actually
numbers by themselves (E.g. 2s= 2*s). Hence, she cannot add them together to get the digit (6) as her prefix. It should have been: 2s+4t= 2s+4t

1. The exception is that when s=t=1. It would be:
(2*1)+(4*1)=6*1*1
=
2+4=6

This shows that this expression, 2s+4t, is correct in this scenario.

2. This expression is also correct when s=t=0. It would be:
(2*0)+(4*0)=6*0*0
=
0+0=0

This shows that this expression is correct when placed in this scenario.

3. Let a=4 and b=5,
3a+b=3*a+b
3ab=3*a*b
So,
3a+b=3*4+5=17
3ab=3*4*5=60
As you can see, the numbers are vastly different so she is wrong.
The correct way should be 3a+b=(3a+b)

Let s=4 and let t=5,
2s+4t=2*4+4*5=28
6st=6*4*5=120
So, she is again incorrect.
Thus, the correct way she should do it is 2s+4t=(2s+4t) as s and t is 2 different unknown variable and cannot be added together.

1. Her statement is only correct when a=0 and b=0

2. As for the second question, she is still only right when s=0 and t=0

3. Apart from 0, if s=1 and t=1 her equation is correct

4. 1) Jane is wrong. a + b = a + b, not = ab as this shows a * b. Thus, the expression should be witten/remain as 3a+b.

2) Jane is wrong. 2s + 4t is not 6st as this st only occurs when there is multiplication and adding of the 2 and 4 to get 6 only applies if it is only adding of the digits, without the alphabets beside the numbers. Thus, the expression should be left as 2s + 4t or, 2(s + t) for factorisation.

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6. Only when s=0 and t=0, Jane's statement will be correct.

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8. Under the scenario s=1 t=1, or s=0 or t=0

9. Jane is wrong as the terms a and b is not the same and cannot be added that way.
She is correct only when a and b is 0.
Jane is wrong as the terms s and t is not the same and cannot be added that way.
She is correct only when s and t is 0.

10. 1) For this equation, it is false, with an exception of if a = 0 and b = 0.
2) For this equation, it is false, with an exception of if a = 0 and b = 0, and if a = 1 and b = 1

11. 1)Jane is only correct when a and b is 0.
2)Jane is only correct when s and t is 0.

12. >>2s+4t=6st<<

(2*s)+(4*t)
=(2+4)*(s+t)
=6*(s+t)

s+t≠s*t

unless s=0 and t=0

13. Jane is wrong for both.

1) 3a+b = 3a+b = (3 * a) + b but not 3ab as 3ab is 3a * b not (3 * a) + b.
2) 2s+4t = 2s+4t = (2 * s) + (4 * t) but not 6st as 6st is 6 * s * t but not (2 * s) + (4 * t).

1. She is correct only if a , b , s and t = 0.

2. Or s and t = 1

14. If s and t = 1, Jane will be correct as 2s=2 , 4t=4, so 2s + 4t =6st

1. It applies to s and t = 0 too.

15. s = 1 and t = 1

16. No Jane is wrong for both questions.
Only when s=0 and t=0

1. or both sides are the same

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18. Jane is only right if 'a' and 'b' are both the unknown are zero or one.
This is the same for both scenarios. This would be wrong if the numbers are not zero or one and other numbers

19. 2) 2s+4t is not 6st because it is not multiplication but addition. but her statement is correct when s and t is both 0 or 1

20. She is correct only when s=0 and t=0 or s=1 and t=1

21. 2) Jane is also correct when s and t is 1.

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1. Sorry refer to the second post.

23. jane is wrong.
For a) a+b is not ab as it is not axb. Example: Take a as 3 and b as 4 . If it is a+b=4+3=7 then4x3=12. Then 3ab is 3x3x4=36, and 3a+b is 9+4=13. So 3a+b is not 3ab
b) if you take s as 3 and t as 4. Then 6st would 6x3x4=72, while 2s+4t is 6+16=22. So 2s+4t is not 6st.

SO HOW TO MAKE 6st THE SAME???

eg s and t is 0,1
0 as 6x0x0=0
1 as 6x1x1=6

24. Jane is wrong for both the equations. 3a+b is equal to (3a+b) as the operation in this equation is not *. It is +. Only if the operation in this equation is *. Then the answer will be 3ab. In the second equation the operation again is + not *. Only if the operation is *, then the answer will be 6st. This reasoning is applicable to the second equation unless the value of a,b,s and t is 0 or 1.

25. 2s + 4t = 6st

E.G If s=2 and t=5

2s=2 x 2 =4
4t= 4 x 5 = 20
20 + 4 = 24(2s + 4t)

BUT...........

6st=(6 x 2 x 5 ) = 60

Thus,2s + 4t does not equal to 6st