divide 180 into two parts such that one-3rd to one part may exceed one-6th of the other part by 6 ?
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let one part be x and other part be (180-x)
then
x/3 = (180-x)/6 +6
x/3 - (180-x)/6 = 6
(2x - 180 + x)/6 =6
3x -180 = 36
x= 72
therefore one part is 72 and other is 108
u can verify it
then
x/3 = (180-x)/6 +6
x/3 - (180-x)/6 = 6
(2x - 180 + x)/6 =6
3x -180 = 36
x= 72
therefore one part is 72 and other is 108
u can verify it
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