It is given that the original car has an initial value of 20 000 dollars and the value decreases by 1500 dollars each year. If we let x represent the number of years that passes and y represent the value of the car, we can write an equation to describe the situation algebraically.
y=20 000-1500x
To determine the value of the original car after 5 years, we can substitute x=5 into the equation and solve for y.
The value of the car at year 5 will be 12 500 dollars.
Different Car
For the different car to have the same value as the original car in 5 years, it must also be worth 12 500 dollars. Let's use a linear equation in slope-intercept form to represent the value of the different car.
y=mx+b
In this form b is the initial value and m is the rate per year by which the value decreases. We will assume that (5,12 500) is a solution to the equation. That will ensure this car meets the requirements of the exercise. Let's begin by substituting (5,12 500) into the equation for (x,y).
Now we can arbitrarily choose the initial value of the car. Suppose this car was initially worth 18 000 dollars. Let's substitute b=18 000 into our equation and solve for m.
If this car is initially worth 18 000 dollars, it must decrease by 1100 dollars each year to reach 12 500 after 5 years. We can now complete the equation.
y=- 1100x+18 000
Let's graph both equations to ensure that both cars are worth 12 500 dollars after 5 years.
Since both lines pass through the point (5,12 500), we can conclude that both cars are worth 12 500 dollars in year 5.
