- Define initial quantity \(x\) and price \(y\) with \(xy = 150\).
- Set up the equation \((x - 5)(y + 5) = 150\) to reflect the change in quantity and price.
- Solve the resulting quadratic equation \(x^2 - 5x - 150 = 0\) to find the initial quantity of \(x=15\).
- Calculate the initial price \(y = 150 / 15 = 10\). The price of each watermelon is \(\boxed{\$10}\).
Explanation
- Analyzing the Problem and Variables
Let \(x\) be the number of watermelons purchased, and let \(y\) be the initial price of each watermelon in dollars. We are given that the total cost for the watermelons was \(\$150\), which gives us the relationship:
\[xy = 150 \implies y = \frac{150}{x}.\]
We are also told that if each watermelon had cost \(\$5\) more (i.e., \(\$y + 5\)), Jane could have bought \(5\) fewer watermelons (i.e., \(x - 5\)). The total cost remains \(\$150\), so:
\[(x - 5)(y + 5) = 150.\]
- Setting up and Simplifying the Equation
Substitute \(y = \frac{150}{x}\) into the equation \((x - 5)(y + 5) = 150\):
\[(x - 5)\left(\frac{150}{x} + 5\right) = 150.\]
Expand the product:
\[x \cdot \frac{150}{x} + 5x - \frac{750}{x} - 25 = 150,\]
\[150 + 5x - \frac{750}{x} - 25 = 150.\]
Subtract 150 from both sides:
\[5x - \frac{750}{x} - 25 = 0.\]
- Solving for the Number of Watermelons
To eliminate the fraction, multiply the entire equation by \(x\):
\[5x^2 - 750 - 25x = 0.\]
Divide by 5 to simplify:
\[x^2 - 5x - 150 = 0.\]
This is a quadratic equation. We can solve it using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-150)}}{2(1)} = \frac{5 \pm \sqrt{25 + 600}}{2} = \frac{5 \pm \sqrt{625}}{2} = \frac{5 \pm 25}{2}.\]
Since the number of watermelons must be positive, we take the positive root:
\[x = \frac{5 + 25}{2} = \frac{30}{2} = 15.\]
- Finalizing the Calculation
Now that we have the initial number of watermelons \(x = 15\), we can find the initial price per watermelon \(y\) using the first relationship:
\[y = \frac{150}{x} = \frac{150}{15} = 10.\]
The initial price of each watermelon was \(\$10\). We can check this: at \(\$10\) per watermelon, 15 watermelons cost \(\$150\). If each cost \(\$5\) more (\(\$15\)), Jane could buy \(150 / 15 = 10\) watermelons, which is exactly \(15 - 5 = 10\) watermelons. The result is verified. The price of each watermelon is \(\boxed{\$10}\).
Examples
This type of problem is essential for understanding resource allocation and cost optimization. For example, a business owner needs to determine how many units of a product to order at a certain price to maintain a specific budget while accounting for potential price changes. By setting up algebraic relationships between unit price and quantity, they can calculate the impact of market fluctuations and make informed purchasing decisions that keep them within their financial limits.