Answer: 3/4 I’m pretty sure. I looked at it and got confused so not 100% sure.
If you are looking for the expanded version, this is it. It is also possible to group either an a from (a^2 + 2b) or a b from (2ab + b^2). Hope this helps!
Answer:


Step-by-step explanation:
Given

Solving (a): Write as inverse function

Represent a(d) as y

Swap positions of d and y

Make y the subject


Replace y with a'(d)

Prove that a(d) and a'(d) are inverse functions
and 
To do this, we prove that:

Solving for 

Substitute
for d in 




Solving for: 

Substitute 5d - 3 for d in 

Add fractions



Hence:
