**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.*

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)

ReplyDeleteNo.

ReplyDelete1. (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

The exception is that when s=t=1. It would be:

Delete(2*1)+(4*1)=6*1*1

=

2+4=6

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

This expression is also correct when s=t=0. It would be:

Delete(2*0)+(4*0)=6*0*0

=

0+0=0

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

Let a=4 and b=5,

ReplyDelete3a+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.

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

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

DeleteApart from 0, if s=1 and t=1 her equation is correct

Delete1) Jane is wrong. a + b = a + b, not = ab as this shows a * b. Thus, the expression should be witten/remain as 3a+b.

ReplyDelete2) 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.

2(s + 2t) instead

DeleteThis comment has been removed by the author.

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

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteUnder the scenario s=1 t=1, or s=0 or t=0

ReplyDeleteJane is wrong as the terms a and b is not the same and cannot be added that way.

ReplyDeleteShe 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.

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

ReplyDelete2) For this equation, it is false, with an exception of if a = 0 and b = 0, and if a = 1 and b = 1

1)Jane is only correct when a and b is 0.

ReplyDelete2)Jane is only correct when s and t is 0.

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

ReplyDelete(2*s)+(4*t)

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

=6*(s+t)

s+t≠s*t

unless s=0 and t=0

Jane is wrong for both.

ReplyDelete1) 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).

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

DeleteOr s and t = 1

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

ReplyDeleteIt applies to s and t = 0 too.

Deletes = 1 and t = 1

ReplyDeleteNo Jane is wrong for both questions.

ReplyDeleteOnly when s=0 and t=0

or both sides are the same

DeleteThis comment has been removed by the author.

ReplyDeleteJane is only right if 'a' and 'b' are both the unknown are zero or one.

ReplyDeleteThis is the same for both scenarios. This would be wrong if the numbers are not zero or one and other numbers

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

ReplyDeleteShe is correct only when s=0 and t=0 or s=1 and t=1

ReplyDelete2) Jane is also correct when s and t is 1.

ReplyDeleteThis comment has been removed by the author.

ReplyDeleteSorry refer to the second post.

Deletejane is wrong.

ReplyDeleteFor 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

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.

ReplyDelete2s + 4t = 6st

ReplyDeleteE.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