Raymond B. answered • 03/17/21

Math, microeconomics or criminal justice

7A +5S =110 multiply by 2 to get

14A+10S =220 subtract

14A+9S = 212 to get

S =$8 for a student ticket

7A=110-5S = 110-40 = 70

A = 70/7 = $10 for an adult ticket

Giovanni Y.

asked • 03/17/21Rob's school is selling tickets to the annual talent show. On the first day of ticket sales the school sold 14 adult tickets and 9 student tickets for a total of $212. The school took in $110 on the second day by selling 7 adult tickets and 5 student tickets. What is the price each of one adult ticket and one student ticket?

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Raymond B. answered • 03/17/21

Tutor

5
(2)
Math, microeconomics or criminal justice

7A +5S =110 multiply by 2 to get

14A+10S =220 subtract

14A+9S = 212 to get

S =$8 for a student ticket

7A=110-5S = 110-40 = 70

A = 70/7 = $10 for an adult ticket

Very common type of question. Two variables in two equations is straightforward to solve. There are many ways to do so, from elimination and substitution to Cramer's Rule. Here elimination is simple.

14a + 9s=212

7a + 5s = 110

Multiply the second equation by -2

14a + 9s = 212

-14a - 10s = -220

Add together and get -1s = -8, so a student ticket is $8

Put it into the second (or first if you want to) and solve for a.

7a + 5(8) = 110

7a + 40= 110

7a = 70

a = 10

Adult tickets are $10. ■

Jon S. answered • 03/17/21

Tutor

5
(6)
Patient and Knowledgeable Math and English Tutor

x = price of adult ticket

y = price of student ticket

14x + 9y = 212

7x + 5y = 110

multiply 2nd equation by 2 and subtract from 1st equation

14x + 9y = 212

-(14x + 10y = 220)

-y = -8

y = 8

substitute 8 for y in 2nd equation and solve for x:

7x + 5(8) = 110

7x = 70

x = 10

$10 for adult ticket, $8 for student ticket

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