SOLUTION: I have a word problem that I am not sure how to put into an algebraic equation(s): Peter can buy a high-efficiency furnace for $2250 that costs $412 per year to operate. He may

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Question 417859: I have a word problem that I am not sure how to put into an algebraic equation(s):
Peter can buy a high-efficiency furnace for $2250 that costs $412 per year to operate. He may also buy a standard furnace for $1710 that costs $466 per year to operate. Find how long it will take for the two furnaces to have the same cost. If Peter intends to live in his house for 7 years which furnace should he buy? What if he wants to live in his house for 20 years?
My instructor did not cover any word problems even close to this one, so if anyone could help me with this I would appreciate it. It is due Tuesday.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Peter can buy a high-efficiency furnace for $2250 that costs $412 per year to operate.
He may also buy a standard furnace for $1710 that costs $466 per year to operate.
Find how long it will take for the two furnaces to have the same cost.
If Peter intends to live in his house for 7 years which furnace should he buy?
What if he wants to live in his house for 20 years?
:
Let y = no. of years that he will use the furnace
:
Write a cost equation for each furnace:
C(y) = 412y + 2250; (High Efficiency)
and
C(y) = 466y + 1710; (Standard)
:
To find when the costs are equal:
Standard = High Efficiency
466y + 1710 = 412y + 2250
466y - 412y = 2250 - 1710
54y = 540
y = 540%2F54
y = 10 yrs, they will cost the same
:
You can confirm this by substituting 10 for y in both equations
:
You can see if he only lives there 7 yrs, the standard would be cheaper.
However, in 20 yrs the high efficiency furnace would make sense
:
:
prove this to yourself, calculate the total cost when y=7 for both furnaces
do the same for y=20.