GSCE FURTHER MATHS QUESTION-DISAGREEMENT WITH MARK SCHEME

So i was helping a friend with this GCSE Further Maths Paper, and there was a one mark question which read:

Bag A has 7x counters
Bag B has 2x counters

5 counters from bag A are taken out and put into bag B. Give an expression for the number of counters in bag B.

I know i asked this before but i needed to understand why the answer is NOT: 2x + (5/7x)
because there are 2x already in bag B and 5 out of what is in bag A(7x) is added into it.

Why is it 2x +5?? <<
on a serious note im confused

B has 2x counters. 5 are taken out of bag A, so you have the 2x originally, + the 5 from bag A.

the expression 7x means 7 times a number
2x means 2 times that same number

So, let's give the example x = 10
then there are 7x = 7(10) = 70 counters in bag A
there are 2(10) = 20 counters in bag B

If you take 5 from bag A, then bag A had 70 - 5 = 65
Bag B now has those 5 counters, so it has 20 + 5 = 25 counters. This is the original (20 = 2x) + the 5 counters that moved.

If you use a different number for x, such as x = 8, you would have
Bag A = 7(8) = 56
Bag B = 2(8) = 16
taking 5 from bag A would leave bag a with 56 - 5 = 51
giving the 5 to bag B means bag B has 16 + 5 = 21
the 16 is the original amount = 2(8) = 2x + 5 that moved

No matter what the common ratio value you use, the bag B only gains 5 counters, so you add 5 to whatever the expression of bag B
***EDIT**
The total number of counters in the bag is "2x" or "7x". It is just a number. It could be 20 and 70, 2 and 7, 6 and 21, or 200 and 700. We don't know the common factor.

Taking out 5, is just taking out the 5 counters, not 5 times the common factor. So we are adding 5 counters to the other bag, not 5x, which would be 5 times the common factor. For example, if the common factor is 10, then the bag of 7x would have 70 counters. We do not remove 5x = 50 counters, we only remove 5. So there is no need to show another variable.