We will form the equations for this problem:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
z = ? Monthly administration fee is notated with z, and that is the this problem's question.
Number of kilowatt hours of electricity used are numbers 1100 and 1500 respectively.
Cost per kilowatt hour is notated with y, but its value is not asked in this math problem, but we can calculate it anyway.
The problem becomes two equations with two unknowns, it is a system, and can be solved with method of replacement:
(1) 1100*y + z = 113
(2) 1500*y + z = 153
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(1) z = 113 - 1100*y [insert value of z (right side) into (2) equation instead of z]:
(2) 1500*y + (113 - 1100*y) = 153
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(1) z = 113 - 1100*y
(2) 1500*y + 113 - 1100*y = 153
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(1) z = 113 - 1100*y
(2) 400*y + 113 = 153
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(1) z = 113 - 1100*y
(2) 400*y = 153 - 113
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(1) z = 113 - 1100*y
(2) 400*y = 40
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(1) z = 113 - 1100*y
(2) y = 40/400
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(1) z = 113 - 1100*y
(2) y = 1/10
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if we insert the obtained value of y into (1) equation, we get the value of z:
(1) z = 113 - 1100*(1/10)
(1) z = 113 - 110
(1) z = 3 dollars is the monthly fee.
credit: freesparrow
