Question

1.3 general linear equations: problem 4
(1 point)
in 2003, the price of a certain automobile was approximately $31,400 with a depreciation of $1,530 per year. after how many years will the cars value be $16,100?
a) 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:
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Explanation:

Step1: Write the linear - equation formula

The general form of a linear equation is $y=mx + b$, where $y$ is the dependent variable, $x$ is the independent variable, $m$ is the slope, and $b$ is the y - intercept. In the context of the car's value, let $V$ be the value of the car and $t$ be the number of years after 2003. The initial value of the car ($t = 0$) is $b=31400$, and the rate of depreciation (slope) is $m=- 1530$. So the equation is $V(t)=-1530t + 31400$.

Step2: Solve for $t$ when $V(t)=16100$

Set $V(t)=16100$ in the equation $V(t)=-1530t + 31400$. We get the equation $16100=-1530t + 31400$.
First, subtract 31400 from both sides:
$16100−31400=-1530t+31400 - 31400$
$-15300=-1530t$
Then divide both sides by - 1530:
$t=\frac{-15300}{-1530}=10$

Answer:

a) $V(t)=-1530t + 31400$
b) 10 years