Could u give the rest of the question?
Let x = # of tickets sold in advance
Let y = # of tickets sold the day of
Cost of the tickets & total sales: 6x & 10y = 6828
You also know y = x + 206
Take the equation mentioned above y = x + 206 and sub it in anywhere the variable y is in the other equations so you'll have this:
6x + 10(x+206) = 6828
Now solve for x to get x = 298
To finish the problem, you must now find the number of y tickets sold.
Sub your x value that you found back into the equation y = x + 206 and you'll get y = 504.
So, 298 tickets were sold in advance and 504 tickets were sold the day of
Answer:
Awww love you man!
(And me lol)
Step-by-step explanation:
Original price: $3,495
Yearly price decrease = 0.2 x 3,495 = $699 per year
Let y be the value of the car and x be the number of years:
y = 3495 - 699x
After 5 years,
y = 3495 - 699(5)
= $0
In order to solve this problem, we set up a linear equation where the variables were the value of the car and the time passed since purchase. The claim that the car will not be anything is true to an extent because its price has dropped to 0 according to our function; however, practically, a car cannot be $0 so it will be worth at least something, however smaller than its original price.
