1. The vowels of the 26 letters of the English alphabet, excluding y, are 5: a, e, i, o and u.
2. The prime numbers less than 10 are 4: 2, 3, 5 and 7.
3. The raffle tickets have one of {a, e, i, o,u} as the first character, and
one of {2, 3, 5, 7} as the second character.
4. Some examples of the raffle tickets: a2, a3, e5, i7 ...
In total there are 5*4=20 possible tickets, as any of the 5 vowels can be combined with any of the 4 primes.
that is, we have a2, a3, a5, a7, e2, e3, ....u7
5.
The only vowel contained in the word "event" is e.
"the digit is the number of letters in the word event":
the word "event" has 5 letters.
-------------------------------------------------------------------------------------------------- Remark: it is not particularly clear whether it is "the number of letters" or "the number of different letters", in which case there would be 4.
For example, if Adam is asked how many letters are there in his name, we expect him to say "4", unless specifically asked "how many different letters are there" --------------------------------------------------------------------------------------------------
The ticket which matches the description can only be e5
6.
P(e5)=(number of ways e5 may occur)/(total number of tickets)= 1/20=0.05
in the top equation, we can get y by itself on one side of the equation:
-x+y=2
y=x+2
now, since we know that y=x+2, in the bottom equation we can substitute x+2 for y (substitution method) and solve:
3x-5y=-4
3x-5(x+2)=-4
3x-5x-10=-4
-2x-10=-4
-2x=6
x=-3
since we now know that x=-3, we can use it to find y by plugging in -3 for x in the top equation (you can do it in either equation, but doing it in the top equation is easier):
