At a carnival, food tickets cost $2 each and ride tickets cost $3 each. A total of $1,240 was collected at the carnival. The number of food tickets sold was 10 less than twice the number of ride tickets sold.The system of equations represents x, the number of food tickets sold, and y, the number of ride tickets sold.
2x + 3y = 1240
x = 2y – 10
How many of each type of ticket were sold?
Answers
The given equations satisfy the given conditions. There are 2 equations and 2 unknowns, so a certain solution can be found.
This can be solved using substitution,
Substituting eqn 2 to eqn 1:
2(2y – 10) + 3y =1240
Simplifying,
y = 180
x = 350
The number of each type of tickets sold are as follows:
number of food ticket sold = 350
number of ride ticket sold = 180
How to solve for the variable in a system of equation?
x = number of food sold
y = number of ride tickets sold
Therefore,
2x + 3y = 1240
x = 2y – 10
Hence,
2x + 3y = 1240
x - 2y = – 10
2x + 3y = 1240
2x - 4y = -20
7y = 1260
y = 1260 / 7
y = 180
Therefore,
x = 2(180) - 10
x = 360 - 10
x = 350
Therefore, 350 food tickets were sold and 180 ride tickets were sold.
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Answers
so find the smallest numbber that al of them will go into
10,6,5
first find th efactors
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so the answer is 2 times 3 times 5=30
bottom number must by 30
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least to greatest the order is 3,10,18
since3/5=6/10 they are equal so the order is
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Answers
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A is on it
answer is A
by sub. by this points in the circle equation
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so the point V lay on the circle
Answers
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Answers
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