Sagot :
We let the price of an adult's ticket be A and C for children.
We have two equations:
4A + 6C = 260...[1]
A + C =55 ... [2]
We need to remove C in order to know the value of A so:
6[2] - [1]
6A + 6C = 330
-4A - 6C = -260
2A = 70
A = 35
Therefore the cost of an adult's ticket is P35
We have two equations:
4A + 6C = 260...[1]
A + C =55 ... [2]
We need to remove C in order to know the value of A so:
6[2] - [1]
6A + 6C = 330
-4A - 6C = -260
2A = 70
A = 35
Therefore the cost of an adult's ticket is P35
Lets represent:
x=cost of adult ticket
y=cost of children ticket
The equation are
4x+6y=260
x+y=55
For the second equation, find the value of x
x+y=55
x=55-y
After finding it, substitute it to the first equation
4(55-y)+6y=260
220-4y+6y=260
Add like terms
220+2y=260
Subtract both sides by 220
2y=40
y=20
Substitute the value of y to any equation
x+y=55
x+20=55
Subtract both sides by 20
x=35
Hope this helps =)
x=cost of adult ticket
y=cost of children ticket
The equation are
4x+6y=260
x+y=55
For the second equation, find the value of x
x+y=55
x=55-y
After finding it, substitute it to the first equation
4(55-y)+6y=260
220-4y+6y=260
Add like terms
220+2y=260
Subtract both sides by 220
2y=40
y=20
Substitute the value of y to any equation
x+y=55
x+20=55
Subtract both sides by 20
x=35
Hope this helps =)