Chapter 5.5, Problem 40E

To determine

The car that will worth the same amount as in car is exactly 5 years.

Answer to Problem 40E

Where the initial value is the y-intercept at $22500 & the rate at which the value decreases is $2000 per year.

Explanation of Solution

Given information:

A car has initial value of $20,000 and decreases in value at a rate of $1500 per year.

Formula used:

  $y=mx+b$

Calculation:

First, derive the equation for the first car. The initial value corresponds to the y-intercept

b = 20000 & the decrease in value rate is the slope m = -1500

Using slope-intercept form, the equation is:

  $\begin{array}{l}y=mx+b\hfill \\ y=-1500x+20000\text{ or}\hfill \\ y=20000-1500x\hfill \end{array}$

At = 5 (5 years), the worth is:

  $\begin{array}{l}y=20000-1500\left(5\right)\hfill \\ y=\$12500\hfill \end{array}$

Therefore, we need to find an equation in the same form that passes through the point (5, 12500) corresponding to the value after 5 years.

Let, the decrease in value he $2000 per year so that the equation is:

  $y=-2000\left(5\right)+b$

Substitute the point to find b:

  $\begin{array}{l}12500=-2000\left(5\right)+b\\ b=22500\end{array}$

The equation for the 2^nd car is:

  $y=22500-2000x$

Where the initial value is the y-intercept at $22500 & the rate at which the value decreases is $2000 per year.

Conclusion:

Where the initial value is the y-intercept at $22500 & the rate at which the value decreases is $2000 per year.