Arthur D. answered 05/17/17
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A+B+C+D=75
A-4=B+4=C/4=4D
A-4=B+4
A=B+8
C/4=4D
C=16D
A+B+C+D=75
(B+8)+B+16D+D=75
2B+17D+8=75
2B+17D=67
in this equation D can be only 1 or 3 because 2B is even and 5 is too big
let D=3
2B+17*3=67
2B+51=67
2B=16
B=8
B+4=12 and 4D=12
therefore A-4=12 and C/4=12
therefore A=16 and C=48
A=16, B=8, C=48, and D=3
48-3=45

Arthur D.
tutor
You're very welcome, John.
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05/18/17
John K.
05/18/17