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Roy and Elisa are buying tickets for the annual school concert. Roy buys 6 adult tickets and 2 child tickets for a total of $66. Elisa buys 5 adult tickets and 4 child tickets for a total of $62. Determine the system of equations that can be used to find the cost of one adult ticket, a, and the cost of one child ticket, c.

User Debbe
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2 Answers

3 votes

Answer:

The system of equations is.....


3a+c=33\\ 5a+4c=62

Explanation:

The cost of one adult ticket is 'a' and the cost of one child ticket is 'c'

Roy buys 6 adult tickets and 2 child tickets for a total of $66.

So, the cost of 6 adult tickets
=6a dollar and the cost of 2 child tickets
=2c dollar.

Thus, the first equation will be:
6a+2c=66 or
3a+c=33 (Dividing both sides by 2)

Elisa buys 5 adult tickets and 4 child tickets for a total of $62.

So, the cost of 5 adult tickets
=5a dollar and the cost of 4 child tickets
=4c dollar.

Thus, the second equation will be:
5a+4c=62

So, the system of equations that can be used to find the cost of one adult ticket and the cost of one child ticket will be.....


3a+c=33\\ 5a+4c=62

User BWA
by
6.5k points
6 votes

Answer:

The system of equations that could be used to find the cost of one adult ticket and one child ticket is:


6a+2c=66\\\\and\\\\5a+4c=62

On solving we get:

Cost of one adult ticket= $ 10

and cost of one child ticket= $ 3

Explanation:

a denote the cost of one adult ticket.

and c denote the cost of one child ticket.

Roy buys 6 adult tickets and 2 child tickets for a total of $66.

i.e. the equation that could be written is:

6a+2c=66-----------(1)

and also Elisa buys 5 adult tickets and 4 child tickets for a total of $62.

This means that the equation that is obtained from this information is:

5a+4c=62--------------(2)

Also, on solving the two equations by the method of elimination

we multiplying equation (1) by 2 and subtract both the equations we get:

12a+4c=132

and 5a+4c=62

-----------------------------------

7a=70

i.e. a=10

and on putting the value of a in equation (1) we get:

60+2c=66

i.e. 2c=66-60

i.e. 2c=6

i.e. c=3

Hence, the cost of one adult ticket=$ 10

and cost of one child ticket= $ 3

User Unlikus
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