Answer:
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Let student tickets be s and adult tickets be a. The number of tickets sold of both adult and student then is s + a = 396. If each student ticket costs $3, then we represent the money equation by tacking the dollar amount onto the ticket. 3s is the cost of one student ticket. 4a is the cost of an adult ticket. The total money from the sales of both is 4a + 3s = 1385. We now have a system of equations we can solve for a and s. If s+a=396, then s = 396-a. We will sub that into the second equation to get 4a + 3(396-a) = 1385. Distributing we have 4a+1188-3a=1385. a = 197. That means there were 197 adult tickets sold. If s + a = 396, then s + 197 = 396 and s = 199. 197 adult tickets and 199 student tickets. There you go!
Answer: The third one
C.
Step-by-step explanation:
Answer:
no solution
Step-by-step explanation:
2x + 4y = 6......reduces to x + 2y = 3
3x = 12 - 6y....reduces to x = 4 - 2y...rearranged is x + 2y = 4
so now we have :
x + 2y = 3
x + 2y = 4
ok....I dont have to go any farther to know that this has no solution because ur equations have the same slope and different y int, this means ur lines are parallel and have no solution because they never cross each others path.
