B=number of ticets sold before
a=number of tickets sold after
cost of a ticket=number of tickets times cost per ticket
beforecost=39.95b
aftercost=54.95a
total cost=925000
39.95b+54.95a=925000
total number tickets=20000
b+a=20000
we have
39.95b+54.95a=925000
b+a=20000
multiply second equation by -39.95 and add to first equatin
39.95b+54.95a=925000
<u>-39.95b-39.95a=-799000 +</u>
0b+15a=126000
15a=126000
divide bot sides by 15
a=8400
sub back
b+a=20000
b+8400=20000
minus 8400 both sides
b=11600
11,600 tickets sold before
8400 tickets sold after
x=1, y=2
Solve the following system:
{y = 5 - 3 x5 x - 4 y = -3
Substitute y = 5 - 3 x into the second equation:
{y = 5 - 3 x5 x - 4 (5 - 3 x) = -3
5 x - 4 (5 - 3 x) = (12 x - 20) + 5 x = 17 x - 20:{y = 5 - 3 x17 x - 20 = -3
In the second equation, look to solve for x:{y = 5 - 3 x17 x - 20 = -3
Add 20 to both sides:{y = 5 - 3 x17 x = 17
Divide both sides by 17:{y = 5 - 3 xx = 1
Substitute x = 1 into the first equation:{y = 2x = 1
Collect results in alphabetical order:Answer: {x = 1 y = 2
Okay, so this is simple. It just needs steps.
Quotient of 18 and 2 = 9.
Sum of 22 and 9 = 31.
31 - 9 = 22.
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
