SOLUTION: helppp
Jen butler has been pricing Speed-Pass train fares for a group trip to New York.
three adults and four children must pay $129.
two adults and three children must pay
Algebra ->
Angles
-> SOLUTION: helppp
Jen butler has been pricing Speed-Pass train fares for a group trip to New York.
three adults and four children must pay $129.
two adults and three children must pay
Log On
Question 915910: helppp
Jen butler has been pricing Speed-Pass train fares for a group trip to New York.
three adults and four children must pay $129.
two adults and three children must pay $92.
Find the price of the adults ticket and the price of a child's ticket.
Adult
Child Answer by Hawksfan(61) (Show Source):
You can put this solution on YOUR website! You can solve this question by either the substitution method or the elimination method. I am going to illustrate the substitution method here.
Lets let a= adult ticket and c = children's ticket
We need two equations for this problem with two unknowns.
The first equation would then be
3a + 4c = 129
The second equation would be
2a + 3c = 92
Let's solve the first equation for a:
3a + 4c = 129
3a = 129 - 4c
a = 129/3 - 4/3c
a = 43-4/3c
Now we can replace a in the second equation:
2a + 3c = 92
2(43 - 4/3c) + 3c= 92
86 -8/3c +3c =92
subtract 86 and covert to thirds
-8/3c + 9/3c =6
1/3c= 6
multiply *3
c = 18
Find a by using one of the 2 equations from the beginning
2a + 3c = 92
2a + 3*18 = 92
2a + 54 = 92
2a = 38
a = 19
check answer with the other equation.
3a + 4c = 129
3*19 + 4*18 =129
57 + 72 =129
129=129