SOLUTION: Separate 71 into two parts such that 40 exceeds ⅔ of one part as much as the other exceeds 16.
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Question 202857: Separate 71 into two parts such that 40 exceeds ⅔ of one part as much as the other exceeds 16.
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
1) x + y =71,,,,,y=71-x
.
2) (2/3 )x -40 = 16-y = 16-(71-x) = -55 +x
.
(-1/3) x = -15
.
x=45,,,,,y=71-45=26
.
check,,,(1) 45 +26 =71,,,ok
.
(2) 2/3 (45)-40 =16-26,,,,,-10 =-10,,,,,ok
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