Ron is selling children's tickets for $3.00 each. I am selling adult tickets for $6.50 each. We had a total income of $125.00 from 30 tickets sold. How many did Ron sell?
To solve this problem, we should identify the information that we know.
- We sold children tickets. We will let the variable C= children's tickets.
- We sold adult tickets. We will let the variable A = adult tickets
- We know we sold 30 tickets
Therefore we will develop an equation for the number of tickets.
C + A = 30 (the total tickets sold)
We also know.
- The children tickets cost $3.00 Therefore we will let $3.00 x C ($3.00C) = amount of money earned from the sale of children tickets.
- The adult tickets cost $6.50 Therefore we will let $6.50 x A ($6.50A) = amount of money earned from the sale of adult tickets.
- We know we earned $125.00 from the sale of all tickets.
Therefore we will develop an equation for the amount of money earned from the sale of tickets.
$3.00C + $6.50A = $125.00
Now we have 2 equations ( a system of equations)
- C+ A = 30
- 3.00C + 6.50A= 125
We will use the elimination method to solve for one of the variables.
We will solve for A by eliminating the variable C.
We will multiply equation 1 by -3
-3C + -3A = -90
We will add the new equation to equation 2.
- -3C + -3A = -90
- 3C + 6.50A = 125
- 0+ 3.50A = 35 (The result of adding the equations)
- We will divide both sides by 3.50
3.50A/3.50 = 35/3.50
A= 10 (10 is the number of adult tickets sold.)
We will substitute 10 into our equation for the number of adult tickets in the 1st equation.
C +10 = 30
We will subtract 10 from both sides
C+10-10 = 30 -10
C= 20 (the number of children tickets sold.)
We will substitute these solutions (A=10, C=20) into our 2nd equation to verify that this is a feasible solution.
- $3.00(20) + $6.50 (10) = $125
- $60 + $65 = $125
- $125 = $125
The solution satisfies the equationn.
Therefore Ron sold 20 tickets.