## create equations to solve: costs and depreciation rates

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### create equations to solve: costs and depreciation rates

The cost of a Frigbox refrigerator is \$950 and it depreciates \$50 each year. The cost of an Arctic Air refrigerator is \$1200 and it depreciates \$100 per year.

1. If a Frigbox and an Arctic Air are bought at the same time, when do the two refrigerators have equal value?
I know I have to make one side equal to the other, but is this using d/t? Would 950 - 50x=1200-100x? and then solve?

2. If both refrigerators continue to depreciate at the same rates, what happens to the values of the refrigerators in 20 years time? What does this mean?
santaclaus

Posts: 41
Joined: Tue Mar 03, 2009 12:17 am

santaclaus wrote:The cost of a Frigbox refrigerator is \$950 and it depreciates \$50 each year. The cost of an Arctic Air refrigerator is \$1200 and it depreciates \$100 per year.

1. If a Frigbox and an Arctic Air are bought at the same time, when do the two refrigerators have equal value?
I know I have to make one side equal to the other, but is this using d/t? Would 950 - 50x=1200-100x? and then solve?

Exactly!

santaclaus wrote:2. If both refrigerators continue to depreciate at the same rates, what happens to the values of the refrigerators in 20 years time? What does this mean?

You have created two expressions for depreciation, and have set them equal. Implicitly, you have created two "depreciation" functions:

. . . . .$\mbox{Frigbox: }\, f_{F}(x)\, =\, 950\, -\, 50x$

. . . . .$\mbox{Arctic Air: }\, f_{AA}(x)\, =\, 1200\, -\, 100x$

Since "x" stands for the number of years, plug "20" in for "x" in each function and see what you get.

stapel_eliz

Posts: 1715
Joined: Mon Dec 08, 2008 4:22 pm