Question

Solve. Assume the exercise describes a linear relationship. When writing a linear equation, write the equation in slope-intercept form.The average value of a certain type of automobile was $13,980 in 1995 and depreciated to $7860 in 1999. Let y be the average value of the automobile in the year x, where x = 0 represents 1995. Write a linear equation that models the value of the automobile in terms of the year x.

Answer 1

The general slope-intercept form is:

y = ax + b

where a is the slope and b is the y-intercept.

The value of b represents the initial value of y, i.e., when x = 0. So for this problem, we have:

b = 13980

since this was the value of y (average value of the automobile) when x = 0 (the year 1995).

Until now, we obtained the equation:

y = ax + 13980

Now, in order to find a, we can use the fact that, for x = 4 (1999 - 1995 = 4), y = 7860:

7860 = a(4) + 13980

Then, we need to isolate the constant a at one side of the equation and find its value:

7860 = a(4) + 13980

7860 = 4a + 13980

7860 - 13980 = 4a

-6120 = 4a

-6120/4 = a

a = -1530

Therefore, the linear equation is:

y = -1530x + 13980