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.

Question

2 years ago

11

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.

Mathematics

2 answers:

motikmotik · Answer 1 · 2022-09-13T12:21:18+03:00

Answer:

<em>The system of equations is.....</em>

Step-by-step 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, <u>the first equation will be</u>: $6a+2c=66$ or $3a+c=33$ <em>(Dividing both sides by 2)</em>

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, <u>the second equation will be</u>: $5a+4c=62$

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.....

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

padilas [110] · Answer 2 · 2022-09-13T14:07:27+03:00

<h2>Answer:</h2>

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

<h2>Step-by-step explanation:</h2>

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