Suppose a car with an initial value of $25400 depreciates by $820 every year. Set up an equation that answers the the question: How long will it take for the value of the car to decrease to $7500? Use y as your variable.

To find out how long it will take for the value of the car to decrease to $7500, we need to set up a linear equation based on the given depreciation information.

We know:
- The car’s initial value is $25,400.
- The car depreciates by $820 each year.
- We are looking for when the value of the car will be $7,500.

Let $ y $ represent the number of years it takes for the car to depreciate to $7,500.

### Equation setup:
The value of the car decreases by $820 each year, so after $ y $ years, the car’s value is:

\[
\text{Car's value} = 25400 - 820y
\]

We want the car's value to be $7,500, so we set the equation equal to $7,500:

\[
25400 - 820y = 7500
\]

### Solve for $ y $:

1. Subtract 7,500 from both sides:
   \[
   25400 - 7500 = 820y
   \]
   \[
   17900 = 820y
   \]

2. Divide both sides by 820:
   \[
   y = \frac{17900}{820}
   \]
   \[
   y = 21.83
   \]

So, it will take approximately 21.83 years (about 21 years and 10 months) for the car's value to decrease to $7,500.

Would you like more details on how we solved this?

### Related questions:
1. What would be the value of the car after 10 years?
2. How long will it take for the car's value to drop to half of its initial value?
3. How much is the car worth after 5 years?
4. If the annual depreciation increases to $1,000, how many years would it take to reach $7,500?
5. What is the value of the car after 15 years?

**Tip**: Linear depreciation means the car's value decreases by the same amount each year, creating a straightforward way to calculate future values.

Suppose a car with an initial value of $25400 depreciates by $820 every year. Set up an equation that answers the question: How long will it take for the value of the car to decrease to $7500? Use y as your variable.

Learn how to calculate how long it will take for a car's value to decrease from $25,400 to $7,500 with a linear depreciation model of $820 per year. This step-by-step algebra solution is suitable for students in Grades 8-10.

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