I believe the answer might be:
x = 28/19
Steps:
Expand
3(5 - x)/2 - 4(3 + 2x)/5
-3x + 15 - 8x + 12
------------ -----------
2 5
Find the LCD for the equation above:
(15-3x) * 5 - (12+8x) * 2
------------- --------------
10 10
Now we combine them since the denominators are equal:
5(-3+15) - 2(8x+12)
--------------------------
10
We expand above:
-31x+51
-----------
10
2x^2+3
x=3
2(3)^2+3
2(9)+3
18+3
21 is the answer.
Given: 

A.)Consider





Also,





Since, 
Therefore, both functions are inverses of each other.
B.
For the Composition function 
Since, the function
is not defined for
.
Therefore, the domain is 
For the Composition function 
Since, the function
is not defined for
.
Therefore, the domain is 