Answer:
Step-by-step explanation:
x-y=7
-3x+9y=-39
Divide the second equation by 3
-x +3y = -13
Add this to the first equation
x-y=7
-x +3y = -13
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0x +2y = -6
Divide by 2
2y/2 = -6/2
y = -3
Now find x
x-y = 7
x -(-3) = 7
x+3 = 7
Subtract 3 from each side
x = 4
(4,-3)
Or by substitution
x-y=7
solve for x
x = 7+y
-3x+9y=-39
Substitute y+7 in for x
-3(7+y) +9y = -39
Distribute
-21 -3y +9y = -39
Combine like terms
-21 +6y = -39
Add 21 to each side
6y = -18
Answer:
20 games won
Step-by-step explanation:
duh
It is not reasonable that she has 1/2 of her homework to do. She did 3/6 (1/2) at home. She actually has 1/6 of her homework left to do.
Answer: 894
Step-by-step explanation:
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
------------------------------------------------
(1) z = 113 - 1100*y
(2) 400*y = 153 - 113
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(1) z = 113 - 1100*y
(2) 400*y = 40
------------------------------------------------
(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.
