To answer the question, let x be the cost of each hamburger, y be the cost of each medium fries, and z be the cost of each medium drink. The equations that are described in the problem are, (Miller ) 4x + 3y = 18.69 (James) x + 2y + z = 8.66 (Steven) 2x + y + z = 10.27 Solving simultaneously for the values of the variables give x = 3.36, y = 1.75, and z = 1.8. Thus, each hamburger costs $3.36. Each medium fries cost $1.75, and each drink costs $1.8.
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
