SOLUTION: I am having difficulty with a linear equation word problem.
Band A charges $600 to play for the evening. Band B charges $350 plus $.125 for each ticket sold.
-Write a linear eq
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Band A charges $600 to play for the evening. Band B charges $350 plus $.125 for each ticket sold.
-Write a linear eq
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Question 181756: I am having difficulty with a linear equation word problem.
Band A charges $600 to play for the evening. Band B charges $350 plus $.125 for each ticket sold.
-Write a linear equation for the cost of each band
-Graph each equation and find the number of tickets for which the cost of the two bands would be equal.
I tried to find the two equations and came up with -
y= 5/4x - 280
y= x - 600
Thanks. Answer by Mathtut(3670) (Show Source):
You can put this solution on YOUR website! I am having difficulty with a linear equation word problem.
Band A charges $600 to play for the evening. Band B charges $350 plus $.125 for each ticket sold.
-Write a linear equation for the cost of each band
-Graph each equation and find the number of tickets for which the cost of the two bands would be equal.
I tried to find the two equations and came up with -
y= 5/4x - 280
y= x - 600
:
what I look for first in a word problems is what are they after. first they say they want an equation of the cost....so lets call the cost C. what are the costs made up of in each band. A fixed cost(meaning no variable) and one band charges so much per ticket..... so the number of tickets is unknown...lets call the number of tickets sold t. With that lets write our equations
:
Band A: C=600........straight fixed fee (no variable) C is the cost
Band B: C=350+.125(t)....that is a fixed fee + 12.5 cents per ticket
:
notice how they want you to graph and FIND the number of tickets for which the cost of each band will be equal
:
This is what the want Cost of Band A = Cost of Band B
:
so lets set the equations equal to each other and solve for t
:
600=350+.125t
:
.125t=250
: so Band B needs to sell 2000 tickets for the cost to be equal to band A.
:
here is the graph, Lines will intersect at (2000,600)
hard to dicipher in this graph.
If the price per ticket is not 12.5 cents then substitute in what it should be and make the necessary adjustments
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