SOLUTION: I need help with this word problem. After 8 months of use, the value of Kathy's computer had depreciated to $890. After 10 months, the value had gone down to $750. Assume that

Algebra ->  Linear-equations -> SOLUTION: I need help with this word problem. After 8 months of use, the value of Kathy's computer had depreciated to $890. After 10 months, the value had gone down to $750. Assume that      Log On


   



Question 124751: I need help with this word problem.
After 8 months of use, the value of Kathy's computer had depreciated to $890. After 10 months, the value had gone down to $750. Assume that the value of the computer, V, is a linear function of the number of months of use, t.
a) Write an equation for the value of the computer after t months of use.
b) How much did the computer cost originally?
c) What was the monthly depreciation?

Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
After 8 months of use, the value of Kathy's computer had depreciated to $890. After 10 months, the value had gone down to $750. Assume that the value of the computer, V, is a linear function of the number of months of use, t.
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You have two points:
(8,890) and (10,750)
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a) Write an equation for the value of the computer after t months of use.
Slope = (750-890)/(10-8) = -140/2 = -70
Form is V=mt+b ; you have V, m, t, and can solve for "b".
750 = -70*10 + b
b = 1450
EQUATION:
V(t)=-70t + 1450
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b) How much did the computer cost originally?
V(0) = -70*0+1450 = 1450
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c) What was the monthly depreciation?
$70
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Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
The present value (V) of the computer is equal to the original cost (C) less the depreciation (D).
So you can start by writing:
V+=+C-D
But, you are to assume that the depreciation (D) is a linear function of (directly proportional to) the number of months of use (t). So the depreciation (D) can be expressed as:
D+=+kt where k is the constant of proportionality. You can find the value of k as follows:
First, rewrite the equation for the value V in terms of the months of use, t, by replacing D with kt, as:
a) V+=+C-kt You can write two equations, one for each of the given depreciation values and months of use.
V%5B1%5D+=+C-kt%5B1%5D
V%5B2%5D+=+C-kt%5B2%5D
Now substitute the given values of V%5B1%5D+=+890, t%5B1%5D+=+8, V%5B2%5D+=+750, and t%5B2%5D+=+10 to get the two equations:
1) 890+=+C-k%2A%288%29 Rewrite this as:C+=+890%2B8%2Ak
2) 750+=+C-k%2A%2810%29 Rewrite this as:C+=+750%2B10%2Ak
Now set these two equations equal to each other to get:
890%2B8%2Ak+=+750%2B10%2Ak Simplify and solve for k. Subtract 750 from both sides.
140%2B8%2Ak+=+10%2Ak Subtract 8*k from both sides.
140+=+2%2Ak Divide both sides by 2.
k+=+70
b) Now you can find the original cost of the computer.
C+=+890%2B8%2Ak Substitute k = 70.
C+=+890%2B8%2A70
C+=+890%2B560
C+=+1450 or...
C+=+750%2B10%2Ak
C+=+750%2B10%2A70
C+=+750%2B700
C+=+1450
The original cost of the computer is $1,450.00
The monthly depreciation is just the value, k, or $70.00 per month.
Or you can calculate it as:
%28C-D%5B1%5D%29%2Ft%5B1%5D+=+%281450-890%29%2F8 =560%2F8+=+70 or...
%28C-D%5B2%5D%29%2Ft%5B2%5D+=+%281450-750%29%2F10=700%2F10+=+70