SOLUTION: Susan is writing a linear equation for the cost of her cell phone plan. In the first month, she talks for 52 minutes and is charged $19.41. In the second month, she talks for 380 m

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Susan is writing a linear equation for the cost of her cell phone plan. In the first month, she talks for 52 minutes and is charged $19.41. In the second month, she talks for 380 m      Log On

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Question 1004934: Susan is writing a linear equation for the cost of her cell phone plan. In the first month, she talks for 52 minutes and is charged $19.41. In the second month, she talks for 380 minutes and is charged $45.65.
A) How much money does Susan's cell phone company charge for each minute?
B) Write an equation that Susan can use to determine the cost of her cell phone plan, y, as it relates to the number of minutes used, x.

im so confused

Found 2 solutions by addingup, josmiceli:
Answer by addingup(3677) About Me  (Show Source):
You can put this solution on YOUR website!
She has a fixed service charge every month and then a cost per minute. So her equation will be:
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Total cost= Service Charge + $minute*x
TC= SC+$min*x
--------------------------------------
Please read over the problem and make sure you provide us with all the information. The way the problem is written right now does not provide a point of reference to establish how much the service charge is.
Thanks,
J

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the charges per minute for the 2 months
are not the same, then I can assume there is
a fixed monthly charge also
--------------------------
The linear equation has the form:
+y+=+m%2Ax+%2B+b+
+y+ is the total charge for a particular month
+m+ is the [ cost ] / [ min ] which has to be determined
+b+ is the fixed monthly cost which will be the same
for every month which has to be determined
-------------------------
With the given data, I can say:
(1) +1941+=+m%2A52+%2B+b+
(2) +4565+=+m%2A380+%2B+b+
( both y's are in cents )
------------------------
There are 2 equations and 2 unknowns, so it's solvable
Subtract (1) from (2)
(2) +4565+=+m%2A380+%2B+b+
(1) +-1941+=+-m%2A52+-+b+
------------------------
+2624+=+328m+
+m+=+8+ ( this is [ cents ] / [ min ] )
This is the rate she is being charged
plug this value back into either of the equations to find +b+
(1) +1941+=8%2A52+%2B+b+
(1) +1941+=+416+%2B+b+
(1) +b+=+1525+
----------------
The linear equation is:
+y+=+8x+%2B+1525+ ( in cents )
---------------------------
check:
(2) +4565+=+m%2A380+%2B+b+
(2) +4565+=+8%2A380+%2B+b+
(2) +4565+=+3040+%2B+b+
(2) +b+=+1525+
OK
Here is a plot of the equation:
+graph%28+400%2C+400%2C+-50%2C+500%2C+-600%2C+6000%2C+8x+%2B+1525+%29+