Question

The value of a car decreases at a constant rate. After 3 years, the value of the cat is $15,000. After 2 more years, the value of the car is $11,000. Write and solve a linear equation to find the value of the car after 8 years.

After 8 years how much is the car worth?

Answer 1

The above answer is better

Answer 2

to get the equation of any straight line, we simply need two points off of it, let's use the provided values

`\begin{array} \cline{1-2} \stackrel{x}{years}&\stackrel{y}{value}\\ \cline{1-2} 3&15000\\ 5&11000\\ \cline{1-2} \end{array}\hspace{5em} (\stackrel{x_1}{3}~,~\stackrel{y_1}{15000})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{11000})`

`\stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{11000}-\stackrel{y1}{15000}}}{\underset{\textit{\large run}} {\underset{x_2}{5}-\underset{x_1}{3}}} \implies \cfrac{ -4000 }{ 2 } \implies - 2000 \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{15000}=\stackrel{m}{- 2000}(x-\stackrel{x_1}{3})`

`y-15000=-2000x+6000\implies {\Large \begin{array}{llll} y=-2000x+21000 \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{after 8 years, x = 8}}{y = -2000(8)+21000}\implies \boxed{y=5000}`