Respuesta :
a = 2s
c + 2s = 17 .....c + a = 17 (I subbed in "a" for 2s)
2c + a = 24
c + a = 17 .....(-1)
2c + a = 24
---------------
-c -a = -17...result of multiplying by -1
2c + a = 24
---------------add
c = 7
2c + a = 24
2(7) + a = 24
14 + a = 24
a = 24 - 14
a = 10
a = 2s
10 = 2s
10/2 = s
5 = s
so, senior tickets (s) cost $ 5, adult tickets (a) cost $ 10, and child tickets (c) cost $ 7
1 adult ticket + 1 senior ticket would cost $ 15
c + 2s = 17 .....c + a = 17 (I subbed in "a" for 2s)
2c + a = 24
c + a = 17 .....(-1)
2c + a = 24
---------------
-c -a = -17...result of multiplying by -1
2c + a = 24
---------------add
c = 7
2c + a = 24
2(7) + a = 24
14 + a = 24
a = 24 - 14
a = 10
a = 2s
10 = 2s
10/2 = s
5 = s
so, senior tickets (s) cost $ 5, adult tickets (a) cost $ 10, and child tickets (c) cost $ 7
1 adult ticket + 1 senior ticket would cost $ 15
The senior tickets (z) cost $ 5, adult tickets (y) cost $ 10, and child tickets (x) cost $ 7.
It cost $15 to buy one adult ticket and one senior ticket.
Let the cost of the children, adults and seniors ticket be $x, $y and $z respectively. Then according to question
[tex]y= 2z ...(1)\\x+ 2z = 17 \\x + y = 17...(2)[/tex] [ By substituting [tex]2z=y[/tex]]
[tex]2x + y= 24 ...(3)[/tex]
Subtracting equation (2) and (3), we get
[tex]x=7[/tex]
Put [tex]x=7[/tex] in (2), we get
[tex]y=17-7\\y=10[/tex]
Put [tex]y=10[/tex] in (1), we get
[tex]10=2z\\z=5[/tex]
Therefore, senior tickets (z) cost $ 5, adult tickets (y) cost $ 10, and child tickets (x) cost $ 7.
Now, cost of 1 adult ticket + 1 senior ticket
[tex]=10+5\\=15[/tex]dollar
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