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.
Since none of the terms have the same variables the like terms would be (A) because they are both constants
Use the distance formula
And just plug in your numbers and solve
We can write a system of equations:
1x + 10y = 182
x + y = 56
Where 'x' is the number of $1 bills, and 'y' is the number of $10 bills.
To find this we can solve using substitution.
Re-arrange the 2nd equation:
x + y = 56
Subtract 'y' to both sides:
x = -y + 56
Now we can plug in '-y + 56' for 'x' in the first equation.
1x + 10y = 182
1(-y + 56) + 10y = 182
-y + 56 + 10y = 182
Subtract 56 to both sides:
-y + 10y = 126
Combine like terms:
9y = 126
Divide 9 to both sides:
y = 14
Now we can plug this into any of the two equations to find the 'x' value.
x + y = 56
x + 14 = 56
Subtract 14 to both sides:
x = 42
So our final answer is (42, 14).
This means that the motel clerk had 42 $1 bills, and 14 $10 bills.
Answer: 12-2r
Step-by-step explanation:
3r+12-5r
= -2r+12 or 12-2r