The lest common multiple of 19 and 14 is 266
hope this helped
Set x as adult tickets.
Set y as children's tickets.
x + y = 15
30x + 20y = 270
Solve for x in the first equation.
x + y = 15
x = 15 - y
Plug this into the second equation.
30x + 20y = 270
30(15 - y) + 20y = 270
450 - 30y + 20y = 270
450 - 10y = 270
-10y = -180
y = 18
If there is 18 childrens tickets, there should be -3 adult tickets.
This is impossible, and this impossible answer occured because the question is written wrong.
There are a total of 15 tickets
The smallest costing ticket is the childrens ticket, which costs 20$.
If he only bought children tickets, this would be 20x15 which is 300$.
300$ is over 270$, which makes the question impossible.
Less than or equal to 121 adult tickets.
You would do 1180.10/9.70
Which equals 121
Since we don't know how many children tickets were, sold it could be less than or equal to 121.
To find the biggest or close number say 70 is close to 75 or 70 the answer would be 70 I think but not sure
