In 2003, the price of a certain automobile was approximately $32,400 with a depreciation of $2,050 per year. After how many years will the car's value be $7,800?

Question

) Write an equation to model the problem. Let t represent the number of years after 2003. For example, the year 2005 would be represented by t = 2.

Answer: 

b) Solve the equation to find the answer to the question above. (Note: Include the units, in this case years.)

Answer: 

Answer 1

If you know how to do the equation of a line, y=mx+b, you're ready to work this problem, except we'll be using t instead of x.

The slope, m: A slope is a rate of change. In this problem they tell us that the rate of change is a decrease of $2,050 per year. That's a negative rate of change, so m = -2,050.

The y intercept, b: The y-intercept is the starting value. In this problem, they tell us that the value of the car was $32,400 at the start. I know, they don't say "at the start," they say "in the year 2003" (wow, that was a long time ago!). But hey, we just say time starts for us in 2003, that's our year 0. So b = 32,400.

Plug in the slope and the y intercept: y = -2,050 t + 32,400

That's your equation. For example, if t=2 (after 2 years or in 2005), y = -2,050 (2) +32,400 = 28,300. In the year 2005, the car's value has sunk to $28,300.

To find out when the car's value will be $7,800, we let that be our y:

7,800 = -2,050 t + 32,400

Then use algebra to solve for 2. Try that out yourself and if you're still stuck, leave another message. Good luck.

Answer 2

32400-2050= 3400(1-r)^1

3150/32400= 1-r

r= 2050/32400 =205/3240

r= 41/648= about 6.33% annual deprectiation

7800 = 32400(1-2050/3400)^t= 3400(41/680)^t

.24 = (41/648)^t

.24 = .633^t

t= 3.1years = 2003+3.1

Feb. 2007

but you probably want 2015

initially 32400

one year 29250

2 years 27200

3 years 25150

t years 32,400-t(2050)=7800

2050t=24600

t= 12 years

2003+12 = 2015

which involves a changing, variable annual rate of depreciation

gernerally the annual rate doesn't change, in nature, science, or finance or rusty cars

although driving the new car off the lot has a jump in depreciation, immediately.