Need help with part c of this AL Maths Question

Here is the solution with the question. https://www.quora.com/profile/Bravewarrior/Need-help-with-part-c

How would I get to 0.01 being the value to substitute into the expansion? Thanks in adavnce!

You want to find 101/103 using (1+x)/(1+3x).
Can you not just sub x=0.01 in and see if / why it works?

Helloo I did sub in 0.01 but my question is how am I supposed to get to 0.01 in the first place?

For other problems like this, how do they work out the x to use? What properties should it have and how do they apply here?

For other problems like this, how do they work out the x to use? What properties should it have and how do they apply here?

icl i never learnt why they sub in 'xyz' values into x but from the questions ive done its usually always 0.01 or 0.1

Helloo I did sub in 0.01 but my question is how am I supposed to get to 0.01 in the first place?

To be completely honest this is the first question in this chapter where they ask me to choose a suitable value otherwise the question normally gives the value to sub in from what I've seen so far 😭😅

Thankyouu I will probably just memorise this and hope for the best tbh...😭

Can you not see a link between 0.01 and 101?

Wbf, if youre just subbing numbers into formulae without understanding why youll come unstuck in exam questions.

For this one, theres only one possibility as the numerator and denominator can be multipled (divided) by an arbitrary constant "k" and the fraction remains unchanged. So here
101 = k(1+x)
103 = k(1+3x)
Solving gives x=0.01 and k=100.

This is a bit unusual and normally the choice of x isnt unique, but there should be an "obvious" one. Obvious would mean something like
In practice it could be related to square .... so its not always 0.1 or 0.01, but thinking like that is a decent place to start.

oh i just clocked how to do it lol, hopefully its not incorrect

take the initial equation and equate it to the fraction/value they provide you with:
(1-x)/(1-3x)=101/103
solve to find x
103(1-x)=103(1-3x)
blah blah and then u get 0.01