, 09.11.2020valensanta2005

# Bryan’s monthly electric bill is determined by adding a flat administration fee to the product of the number of kilowatt hours of electricity used and the cost per kilowatt hour. When he uses 1,100 kilowatt hours of electricity, his bill is \$113. When he uses 1,500 kilowatt hours of electricity, his bill is \$153. What is the monthly administration fee? \$ bercishicicorbin

3

Step-by-step explanation: genyjoannerubiera
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

(1) z = 113 - 1100*y [insert value of z (right side) into (2) equation instead of z]:
(2) 1500*y + (113 - 1100*y) = 153

(1) z = 113 - 1100*y
(2) 1500*y + 113 - 1100*y = 153

(1) z = 113 - 1100*y
(2) 400*y + 113 = 153

(1) z = 113 - 1100*y
(2) 400*y = 153 - 113

(1) z = 113 - 1100*y
(2) 400*y = 40

(1) z = 113 - 1100*y
(2) y = 40/400

(1) z = 113 - 1100*y
(2) y = 1/10

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. broomssymphonie
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

(1) z = 113 - 1100*y [insert value of z (right side) into (2) equation instead of z]:
(2) 1500*y + (113 - 1100*y) = 153

(1) z = 113 - 1100*y
(2) 1500*y + 113 - 1100*y = 153

(1) z = 113 - 1100*y
(2) 400*y + 113 = 153

(1) z = 113 - 1100*y
(2) 400*y = 153 - 113

(1) z = 113 - 1100*y
(2) 400*y = 40

(1) z = 113 - 1100*y
(2) y = 40/400

(1) z = 113 - 1100*y
(2) y = 1/10

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. cselder delphinelilly2846

3

Step-by-step explanation:

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