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For a matinee movie, 100 tickets were sold. The cost was $3 for children, $4 for...

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kristenmarieb... | Student, Grade 10 | (Level 1) Valedictorian

Posted April 3, 2013 at 12:05 AM via web

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For a matinee movie, 100 tickets were sold. The cost was $3 for children, $4 for students, $5 for adults, and a total revenue was $422.  The number of adult tickets sold was 10 less than the sum of the child and student tickets.  How many of each ticket were sold? 

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violy | High School Teacher | (Level 3) Assistant Educator

Posted April 3, 2013 at 1:19 AM (Answer #1)

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Let set x = number of tickets for children
y = numbers of tickets for students.
z = number of tickets for adults.
We know that number of tickets for adults sold was 10 less than the sum of the child and student tickets.
So, z = x + y - 10.
We willhave an equation x + y + z = 100 since, the sum of all the tickets sold is 100.
Revenue is cost times number f tickets sold, so, our second equation is:
3x + 4y + 5z = 422
We can replace the z by x + y - 10.

x + y + z = 100 ===> x + y + x + y - 10 = 100

COmbine like terms.
2x + 2y - 10 = 100

Add 10 on both sides, and divide both sides by 2.
x + y = 55.

We replace the z by x + y - 10 on our second equation.
3x + 4y + 5(x + y - 10) = 422
3x + 4y + 5x + 5y - 50 = 422

Combine like terms.
8x + 9y - 50 = 422

Add 50 on both sides.
8x + 9y = 472

Solve for x on x + y = 55. x = 55 - y.
Substitute x = 55- y on second equation.
8(55 - y) + 9y = 472

440 - 8y + 9y = 472

Combine like terms.
y + 440 = 472

Subtract both sides by 440.
y = 32

Solve for x using x + y = 55.

x + 32 = 55
x = 23

Solve for z using x + y + z = 100.
23 + 32 + z = 100
55 + z = 100
z = 45

Hence, number of tickets sold for children is 23, number of tickets sold for tickets is 32, and for adults is 45.

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