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Question 471178: a piece of equipment(S) annd another piece(G) is both used for 6 hours for $90. equipment S is used on another day for 4 hours and equipment G for 8 for $100. What is the hourly cost of each piece of equipment?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! Rate * Time = Cost
Let s equal the rate of equipment S.
Let g equal the rate of equipment G.
Combined equation is:
(s+g)*T = 90
T is equal to Time which is equal to 6, so this equation becomes:
(s+g)*6 = 90
On another day, S is used for 4 hours and G is used for 8 hours for a total of 100.
The equation for that day is:
s*4 + g*8 = 100
you have 2 equations that need to be solved simultaneously to find s and g.
those equations are:
(s+g)*6 = 90
s*4 + g*8 = 100
these equation can be re-written as:
6*(s+g) = 90
4*s + 8*g = 100
these equations can be re-written as:
6*s + 6*g = 90
4*s + 8*g = 100
we will multiply both sides of the first equation by 1.333333333 (1 and 1/3) to get:
8*s + 8*g = 120
4*s + 8*g = 100
we will subtract the second equation from the first to get:
4*s = 20
we will divide both sides of this equation by 4 to get:
s = 5
we will substitute for s in the first original equation to get:
6*s + 6*g = 90 becomes:
6*5 + 6*g = 90 which becomes:
30 + 6*g = 90
subtract 30 from both sides of this equation to get:
6*g = 60
divide both sides of this equation by 6 to get:
g = 10
we now have:
s = 5
g = 10
these should be the costs per hour (rate) of each of equipments S and G.
go back to your original equations and substitute for s and g to see if those equations prove true.
your original equations are:
6*s + 6*g = 90
4*s + 8*g = 100
these equation become:
6*5 + 6*10 = 90
4*5 + 8*10 = 100
which become:
30 + 60 = 90
20 + 80 = 100
both these equations are true, so the values for s and g are good.
the hourly cost for equipment S is 5 dollars per hour.
the hourly cost for equipment G is 10 dollars per hour.
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