SOLUTION: I needed to use a situation where I compared to phone plans then develop a linear algebraic equation. Additionally, I need to determine where the 2 cost options are equivalent and

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Question 486371: I needed to use a situation where I compared to phone plans then develop a linear algebraic equation. Additionally, I need to determine where the 2 cost options are equivalent and then depict the situation graphically. So far I have 2 phone plans. Plan A is $39.99 a month for 600 minutes with $20 for texting and $.55 for each additional minute (needs a total of 750 minutes). Plan B is $49.99 a month for 700 minutes with $10 for texting and $.50 for each additional minute (need a total of 750 minutes). Here's what I have:
y=.55(x-600)+ (39.99+20)
y=.55(750-600)+59.99
y=.55(150)+59.99
y=$142.49 Am I on the right track? Not sure how to find equality and graph it?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
your plans are too difficult to show simply with one linear equation each.

try this.

plan A = $60 per month and $.10 a minute.
plan B = $20 per month and $.20 a minute.

the equation for plan A would be:
y = .10x + 60
the equation for plan B would be:
y = .20x + 20

they will intersect when .10x + 60 = .20x + 20
that occurs at x = 400 minutes.

that occdurs when the total cost for each plan is equal to:
40 + 60 = 100 for plan A.
80 + 20 = 100 for plan B.

the graph of these equation is shown below:

the vertical line is at x = 400 minutes.
the horizontal line is at y = 100 dollars.

plan A is the line that intersects with the y axis at y = 60
plan B is the line that intersects with the y-axis at y = 20

plan B is cheaper when x is smaller than 400 minutes.
plan A is cheaper when x is greater than 400 minutes.