## The value of a car decreases at a constant rate. After 3 years the value of the car is \$15,000. After 2 more years the value of the car is \$

Question

The value of a car decreases at a constant rate. After 3 years the value of the car 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.

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2 weeks 2021-10-13T03:20:30+00:00 1 Answer 0

The Linear equation: V (x) =  21,000 – 2000x

Where V = Value of Car after a time period (\$)

x = Number of years (yrs)

Value of Car after 8 years = \$5,000

Step-by-step explanation:

Let the value of the car and the number of years be related by the Linear equation :

V(x) = Mx +C——————————————– (1)

Where V =Value of Car after a time period (\$)

x  = Number of years (yrs)

M = Slope of the linear relationship

C = The intercept of the straight line on the value axis

From the question two different coordinates of Value and the number of years where given: (V₁ ,x₁)  and (V₂ ,x₂)

First coordinate (V₁ ,x₁)  = (\$15,000 , 3)

Second coordinate (V₂ ,x₂)  = (\$11,000, 5)

These can be substituted into equation (1) and (2) to calculate the for M & C

Substituting the first coordinate into (1) we have :

15,000 =  3M +C—————————————————————–(2)

Substituting the second coordinate into (1) we have :

11,000 = 5M +C——————————————————————(3)

Solving equation  (2) and (3) simultaneously using elimination method, we have:

15,000 =  3M +C

11,000 = 5M +C

4000 = -2M

M = -2000

Substituting the value of M into equation (3) we have:

11,000 = 5M +C

11,000 = 5(-2000) +C

11,000 = -10,000 +C

C =21,000

Substituting the value of M and  C into equation (1), we have the Linear relationship for Value  and the number of years

V (x) = -2000x + 21,000—————————————————————- (4)

V (x) =  21,000 – 2000x —————————————————————- (5)

Substituting x = 8 years into equation (5) we have:

V (x) =  21,000 – 2000x

= 21000 – 2000 (8)

= 21000 – 16000

= \$5000