A local electronics store will sell seven AC adapters and two rechargeable flashlights for $86, or three AC adapters and four rechargeable flashlights for $84. What is the price of a single AC adapter and a single rechargeable flashlight?
7a+2f = 86
3a+4f = 84
a=8
f=15
Let's assume the price of a single AC adapter is 'A' dollars and the price of a single rechargeable flashlight is 'F' dollars.
According to the given information, we can create two equations:
Equation 1: 7A + 2F = 86 (seven AC adapters and two rechargeable flashlights for $86)
Equation 2: 3A + 4F = 84 (three AC adapters and four rechargeable flashlights for $84)
To solve these equations simultaneously, we can use the method of substitution.
Using Equation 2, we can isolate 'A' in terms of 'F':
3A = 84 - 4F
A = (84 - 4F) / 3
Now substitute this value of 'A' into Equation 1:
7((84 - 4F) / 3) + 2F = 86
Multiplying both sides by 3 to eliminate the fraction:
7(84 - 4F) + 6F = 258
Distributing 7 across (84 - 4F):
588 - 28F + 6F = 258
Simplifying:
-22F = 258 - 588
-22F = -330
F = (-330) / -22
F = 15
Now substitute this value of 'F' back into Equation 2 to find 'A':
3A + 4(15) = 84
3A + 60 = 84
3A = 24
A = 24 / 3
A = 8
Therefore, the price of a single AC adapter is $8 and the price of a single rechargeable flashlight is $15.
To find the price of a single AC adapter and a single rechargeable flashlight, we can set up a system of linear equations based on the given information.
Let's assume the price of a single AC adapter is "A" dollars, and the price of a single rechargeable flashlight is "F" dollars.
According to the information given, we can write two equations:
Equation 1: 7A + 2F = 86 (seven AC adapters and two flashlights cost $86)
Equation 2: 3A + 4F = 84 (three AC adapters and four flashlights cost $84)
We now have a system of linear equations. We can solve this system using various methods, such as substitution, elimination, or matrices. In this case, let's use the substitution method.
From Equation 1, we can express F in terms of A:
2F = 86 - 7A
F = (86 - 7A) / 2
Substituting this value of F into Equation 2, we can solve for A:
3A + 4((86 - 7A) / 2) = 84
3A + 172 - 14A = 168
-11A = -4
A = (-4) / (-11)
A = 4/11
So, the price of a single AC adapter is $4/11.
Now, substitute the value of A into Equation 1 to find the price of a single rechargeable flashlight:
7(4/11) + 2F = 86
28/11 + 2F = 86
2F = 86 - (28/11)
2F = 946/11 - 28/11
2F = 918/11
F = (918/11) / 2
Therefore, the price of a single rechargeable flashlight is $918/22, which simplifies to $41/11.