Penny purchases 100 tickets for her youth services group to attend a waterpark. Child admissions are $14.00 each while adult admissions are $19.00 each. If the total cost for the tickets was $1470, how many of each type of ticket did she buy?

 
This question requires you to set up a system of equations.  First, you must identify your variables.  Since we want to know how many adult tickets and how many child tickets were bought, those will be our variables.  So let:
a = # of adult tickets sold and c = # of child tickets sold
The first sentence says, "Penny purchased 100 tickets..."  This will be used to make our first equation.  The total number of both adult and child tickets should be 100, so our first equation is 
a + c = 100
Then it says, "Child admissions are $14 each while adult admissions are $19 each."  And it states that the total cost of the tickets is $1470.  This will be used to make our second equation, which is
19a + 14c = 1470
There are a few ways to solve this system of equations.  We'll go through two of them below.
1) One method to solve a system is by substitution.  You must solve one equation for one of the variables, then substitute that into the second equation.  For this problem, the first equation is very easy to use to solve for a variable since the variables do not have coefficients.  So
a + c = 100  becomes  c = 100 - a
This will be substituted into our second equation like so:
19a + 14(100 - a) = 1470
From here you can solve the equation for a using algebra:
19a + 1400 - 14a = 1470
5a + 1400 = 1470
5a = 70
a = 14
Now this value for a can be substituted into either original equation to find the value of c.  The first equation is, again, a very easy one to use:
(14) + c = 100
c = 86
So, Penny bought 14 adult tickets and 86 child tickets.
 
2) This can also be solved using the elimination method.  In the elimination method, you are adding the two equations together in an effort to make one of the variables cancel out (eliminate) so that you have just one varible to solve at a time.  To make this happen, you must often multiply one (or both) of the equations by a coefficient so that a variable will eliminate.  Once again, the first equation is very useful for this.  We can choose a variable to eliminate, let's say c.  In the second equation, the coefficient of the variable c is 14.  So we will multiply the first equation by -14 in order to make those two cancel.  The second equation will look like this:
-14(a+c=100)
-14a-14c=-1400
Now we will add the two equations together to get a new single equation with just one variable.
(-14a-14c=-1400)
+(19a+14c=1470)
5a = 70
a = 14
Once again, we can take this value of a and substitute it into one of the original equations to find c.  And we will again get the value 86.
3) This system of equations can also be solved graphically. If the variables are changed to x and y, they can be graphed as lines.  The point of intersection between the two lines is the solution.
 
Again, the final answer is that Penny bought 14 adult tickets and 86 child tickets.