Here’s how to solve this problem. If x represents the unit price, x – 3 will represent the reduced unit price. We will let y represent the number of oranges purchased. We can now create two equations (system of equations):

180/x = y

180/(x – 3) = y + 5

The parts on the left will yield the number of oranges purchased. The first equation will get the original number of oranges purchased (y), and the second equation will get the number after the price reduction (y + 5). Using substitution to solve for x, we take 180/x, which equals y, and plug it into the second equation’s y variable. This gets us the following single variable equation:

180/(x – 3) = 180/x + 5

Next, multiply both sides by (x – 3):

180 = [180(x -3)]/x + 5(x – 3)

180 = (180x – 540)/x + 5x – 15

180 = 180 – 540/x + 5x – 15

Then multiply both sides by x (to remove the last x in a denominator):

180x = 180x – 540 + 5x^2 – 15x

Subtract both sides by 180x, and we have the following quadratic equation:

5x^2 – 15x – 540 = 0

Divide both sides by 5:

x^2 – 3x – 108 = 0

Factor:

(x – 12)(x + 9) = 0

This means that x either equals 12 or -9. Since a negative price can be thrown out, we see that the original price, x, must have been $12. That means the new price (i.e. the reduced price, x – 3) must be $9 per orange.

We can verify this by showing that 180/12 = 15 oranges purchased, and 180/9 = 20 oranges purchased, which indeed means 5 more oranges can be bought.

Accordingly, we now see that the new price per orange is indeed $9.00.