Diaz family spend $ 20 if they bought 3 adults tickets and 1 children's ticket
Solution:
Let "a" be the cost of each adult ticket
Let "c" be the cost of each student ticket
The Spencer family bought 3 adult tickets and 2 children's tickets for $23.50
Therefore, we frame a equation as:
[tex]3 \times a + 2 \times c = 23.50[/tex]
3a + 2c = 23.50 --------- eqn 1
The Yang family bought 2 adult tickets and 4 children's tickets for $25
Therefore, we frame a equation as:
[tex]2 \times a + 4 \times c = 25[/tex]
2a + 4c = 25 ----------- eqn 2
Let us solve eqn 1 and eqn 2
Multiply eqn 1 by 2
6a + 4c = 47 ---------- eqn 3
Subtract eqn 2 from eqn 3
6a + 4c = 47
2a + 4c = 25
( - ) --------------
4a = 47 - 25
4a = 22
Divide both sides of equation by 4
a = 5.5
Substitute a = 5.5 in eqn 1
3(5.5) + 2c = 23.50
16.5 + 2c = 23.50
2c = 23.50 - 16.5
2c = 7
Divide both sides of equation by 2
c = 3.5
Thus, cost of each adult ticket = $ 5.5
Cost of each student ticket = $ 3.5
How much would the Diaz family spend if they bought 3 adults tickets and 1 children's ticket?
Total cost = 3a + 1c
Total cost = 3(5.5) + 1(3.5) = 16.5 + 3.5
Total cost = 20
Thus Diaz family spend $ 20
