Answer: ![y = 1,500(0.7)^x\ \text{and }y = 1,500 - 300x](https://tex.z-dn.net/?f=y%20%3D%201%2C500%280.7%29%5Ex%5C%20%5Ctext%7Band%20%7Dy%20%3D%201%2C500%20-%20300x)
Step-by-step explanation:
Given : Ron buys a lawnmower for $1,500. The salesperson says the value will depreciate about 30% per year (r=0.3) over the next few years.
The exponential decay equation is given by :-
, where A is the initial amount , r is rate of decay and x is the time period.
Then , the equation of depreciation for Ron :-
![y=1500(1-0.3)^x=1500(0.7)^x](https://tex.z-dn.net/?f=y%3D1500%281-0.3%29%5Ex%3D1500%280.7%29%5Ex)
However, his neighbor says it is likely to depreciate about $300 per year, which is linear depreciation.
The linear equation is given by :-
, where 'c' is the initial amount and 'a' is the rate of change.
Then equation of depreciation for his neighbor :-
![y=-300x+1500=1500-300x](https://tex.z-dn.net/?f=y%3D-300x%2B1500%3D1500-300x)
Thus , the system could be used to determine when the two depreciation models will give the same value for the lawnmower:-
![y = 1,500(0.7)^x\ \text{and }y = 1,500 - 300x](https://tex.z-dn.net/?f=y%20%3D%201%2C500%280.7%29%5Ex%5C%20%5Ctext%7Band%20%7Dy%20%3D%201%2C500%20-%20300x)