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?
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
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
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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