<h3>
No, the cost of 1 adult ticket cannot be $ 20 and the cost of 1 children ticket is not $ 8.</h3>
Step-by-step explanation:
Let us assume the cots of ticket for 1 adult  = $ x
Let us assume the cots of ticket for 1 children  = $ y
So, the cost of ticket for 3 adults = 3 x ( Cost of 1 adult ticket)  =  3 x
The cost of ticket for 3 children  = 3 x ( Cost of 1 children ticket)  =  3 y
Also, given the combined cost of 3 adult and children ticket is less than $75.
⇒  3 x +  3  y < 75  ... (1)
Similarly,  the cost of ticket for 2 adults = 2 x
The cost of ticket for 4 children   = 4 y
⇒  2 x +  4  y < 62  ... (2)
Now, solving for the values of x and y, we get:
3 x + 3 y < 75 or, x + y < 25   ⇒ x = 25 - y ( substitute in 2)
2 x +  4  y < 62  or, x +  2 y < 31 
⇒   25  - y + 2 y < 31
or, y + 25 < 31
or, y < 6
⇒ x = 25 - 6  = 19
or, x < 19
So, the cost of 1 adult ticket  is x and is less than $19.
The cost of 1 child ticket  is y  and is less than $6.
Hence, by above statement, NO the cost of 1 adult ticket  can not be $ 20 and the cost of 1 children ticket is not $ 8.