Mrs. Rodrigues was selling two kinds of raffle tickets for $1.50 and $3.50. The number of $1.50 tickets sold was 5 more than three times the number of $3.50 tickets sold. She earned a total of $1,607.50. How many $1.50 tickets were sold?

expensive tickets --- x

cheap tickets ------ 3x+5

3.5x + 1.5(3x+5) = 1607.5

solve for x

To solve this problem, we can set up a system of equations.

Let's say x represents the number of $1.50 tickets sold and y represents the number of $3.50 tickets sold.

From the problem, we know that the number of $1.50 tickets sold was 5 more than three times the number of $3.50 tickets sold. This can be written as:

x = 3y + 5 (equation 1)

We also know that Mrs. Rodrigues earned a total of $1,607.50 from selling the tickets. The total amount earned can be calculated by multiplying the price of each ticket by the number of tickets sold and adding them together:

1.50x + 3.50y = 1607.50 (equation 2)

Now, we have a system of two equations:

x = 3y + 5
1.50x + 3.50y = 1607.50

We can solve this system of equations using substitution or elimination.

Let's use substitution method:

From equation 1, we can express x in terms of y:

x = 3y + 5

Substitute this value of x into equation 2:

1.50(3y + 5) + 3.50y = 1607.50

Distribute the 1.50:

4.50y + 7.50 + 3.50y = 1607.50

Combine like terms:

8y + 7.50 = 1607.50

Subtract 7.50 from both sides:

8y = 1600

Divide both sides by 8:

y = 200

Now, we have found the value of y, which represents the number of $3.50 tickets sold.

To find the number of $1.50 tickets sold (x), substitute the value of y into equation 1:

x = 3(200) + 5
x = 600 + 5
x = 605

Therefore, Mrs. Rodrigues sold 605 $1.50 tickets.