Four devices (A, B, C, D) contain a total 225GB of data. - The device A has half as much data as device B. - The device C has five times as much data as device B and device D combined. - The sum of twice the amount of data on device C and four times the amount of data on the device D is equal to the difference between 500GB and the amount of data on device B. Determine the amount of data (in GB) on each device.
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Step-by-step explanation:
A+B+C+D = 225
B = 2A
C = 5(B+D) = 5B + 5D
2C + 4D = 500 - B ---> B + 2C + 4D = 500
Substitutes B=2A into the other 3 equations, thus eliminating B:
3A + C + D = 225
C = 10A + 5D
2A + 2C + 4D = 500
Substitutes C = 10A + 5D into the first and third equation to eliminate C
3A + (10A + 5D) + D = 225
2A + 2(10A + 5D) + 4D = 500
which simplifies to:
13A + 6D = 225 <---- please label this equation ALPHA: we will need it later
22A + 14D = 500
Elimination method finishes it off...
Since the LCD of 6 and 14 is 42,
Multiplies top equation by -7 and bottom by 3:
-91A + -42D = -1575
66A + 42D = 1500
Adding them together:
-25A = -75
So, A = 3
Then B = 2(3) = 6
By equation ALPHA in bold above:
13(3) + 6D = 225
39 + 6D = 225
6D = 225 - 39
6D = 186
D = 31
Plugging into the original first equation
3 + 6 + C + 31 = 225
C + 40 = 225
C = 185