Respuesta :
Answer:
The system of equations is.....
[tex]3a+c=33\\ 5a+4c=62[/tex]
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 [tex]=6a[/tex] dollar and the cost of 2 child tickets [tex]=2c[/tex] dollar.
Thus, the first equation will be: [tex]6a+2c=66[/tex] or [tex]3a+c=33[/tex] (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 [tex]=5a[/tex] dollar and the cost of 4 child tickets [tex]=4c[/tex] dollar.
Thus, the second equation will be: [tex]5a+4c=62[/tex]
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.....
[tex]3a+c=33\\ 5a+4c=62[/tex]
Answer:
The system of equations that could be used to find the cost of one adult ticket and one child ticket is:
[tex]6a+2c=66\\\\and\\\\5a+4c=62[/tex]
On solving we get:
Cost of one adult ticket= $ 10
and cost of one child ticket= $ 3
Step-by-step 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
