Let's start by figuring out what the equations are that the problem is telling us. We know how many tickets of senior citizens and child tickets were sold each day. We also know how much money was brought in each day. So let's set up the equations for each day:
Let x = senior citizen ticket price
Let y = child ticket price
The equations are:
11x + 14y = 211
7x + 7y = 119
Solving the second equation for x (we do this so we can substitute this result into the x in the first equation) - we can choose either variable and either equation to solve to substitute into the other equation. I just picked this equation because it seemed to be easier. (Sometimes it is just obvious which equation should be used to substitute with - especially if it has a coefficient of 1 in front.)
So we get 7x = -7y + 119 (subtract 7y from both sides)
Divide by 7 on both sides, and we get x = -y + 17.
Use this result for x, and place it into the other equation:
11x + 14y = 211
11(-y + 17) + 14y = 211
Distribute the 11 through the parentheses, and we get:
-11y + 187 + 14y = 211
Combine like terms on the left, and we get:
3y + 187 = 211
Subtract 187 from both sides:
3y = 24
And y = 8. So the price of the child ticket is $8.
Substitute y back into either equation (I am going back to the first equation):
11x + 14y = 211
11x + 14(8) = 211
11x + 112 = 211
Subtract 112 from both sides, and we get:
11x = 99
Divide by 11 on both sides, for a result of:
x = 9.
The price of the senior citizen ticket is $9.
We can check this by substituting x and y into either equation and verifying the sum adds up to the answer on the other side.
7(9) + 7(8) =? 119
63 + 56 =? 119
119 = 119
This is true, so the answer is correct.
