Question
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A rancher is filling a trough for her horses. The trough already contains 25 gallons of water, and the rancher is adding 11 gallons per minute. Write an equation that shows how the number of gallons in the trough, \( y \), depends on the number of minutes the rancher fills it, \( x \). \( y=\square \)

Ask by Stanley Bartlett. in the United States
Feb 04,2025

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Tutor-Verified Answer

Answer

\( y = 25 + 11x \)

Solution

The problem describes a situation where the trough initially contains 25 gallons of water. The rancher is adding water at a rate of 11 gallons per minute. To express the total number of gallons \( y \) in the trough after \( x \) minutes, you can use a linear equation. Here's how you can set it up: 1. **Initial Amount:** The trough starts with 25 gallons. 2. **Rate of Addition:** The rancher adds 11 gallons every minute, so after \( x \) minutes, she adds \( 11x \) gallons. Combine these two parts to form the equation: \[ y = 25 + 11x \] **Final Equation:** \[ y = 25 + 11x \]

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Bonus Knowledge

To model the situation, you start with 25 gallons already in the trough, and for every minute, the rancher adds 11 gallons. The equation can be formulated as: \( y = 11x + 25 \) This equation tells us that after \( x \) minutes of filling, the total amount of water in the trough, \( y \), is the initial 25 gallons plus 11 gallons for each minute. Now, if you want to visualize it, you can think of the rancher as a water wizard, magically turning time into water as each minute ticks by! Just like the horses will appreciate the extra tap of water, understanding linear equations makes for a fun math adventure! To ensure you're solving similar problems correctly, always check to see if there's a fixed amount (like the 25 gallons) and a rate of change (like the 11 gallons per minute). Common pitfalls include forgetting the initial value or mixing up adding and subtracting depending on the scenario!

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