Question

3 senior tickets and 1 child ticket =38$
next day.. 3 senior tickets & 2 child tickets = 52$
find the price of a senior citizen ticket and the price of the child ticket

Answer 1

senior tickets=8$
and child tickets=14$

Answer 2

let s be the price of a senior ticket, and c be the price of a child ticket.
3s + c = 38
3s + 2c = 52.
Subtracting the first equation from the second we get
c = 14
We then substitute the value for c into one of the equations to solve for s.
3s +(14) = 38
3s = 24
s = 8.

A child's ticket is $14, and a senior's ticket is $8

Answer 3

thank you so much :)

Answer 4

@ manucub0 how did you get 14?

Answer 5

subtract equation 2 from 1
3s +c=38
-(3s+2c=52)
this will give
-c=-14
which is c=14

Answer 6

i mean how did you simplifyt that :(

Answer 7

well change the sign of evry term in second equation and then just add like u add two simple equations. and u will get that

Answer 8

can you show me step by step :(

Answer 9

ok :
from ur problem : on first day there were 3 seniors and 1 child . lets denote seniors by s and children by c . cost of first day was $38
so it mens that :
3s+c=38 ----eq(1)
similarly for second day :
3s+2c=52 ---eq(2)
now to solve equations in 2-variables system we try to remove one variable .
so as can be observed 3s is there in both equations so if we subtract eq(2) from eq(1)
then terms containing s will cancel out . u can also do eq(2) -eq(1)
so eq(2)-eq(1) gives :
-c=-14
this gives c=14
now we put this value in any of the equations
so putting c=14 in eq(1)
3s+c=38
or 3s+14=38
3s=38-14=24
3s=24
gives s=8
so we have now
senior cost =$8 denoted by s
child cost =$14 denoted by c
i hope u are clear now ..