SOLUTION: the tens digit of a two number is three greater than the ones digit. the difference between twice the original number and half the number obtained after reversing the digits is 108

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Question 490788: the tens digit of a two number is three greater than the ones digit. the difference between twice the original number and half the number obtained after reversing the digits is 108
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
the tens digit of a two number is three greater than the ones digit. the difference between twice the original number and half the number obtained after reversing the digits is 108
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Let the number be 10t+u
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Equations:
t = u + 3
2(10t+u)-(1/2)(10u+t) = 108
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Modify:
t = u+3
4(10t+u) - (10u+t) = 216
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39t - 6u = 216
t = u+3
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Substitute and solve for "u":
39(u+3) - 6u = 216
33u = 216-3*39
33u = 99
u = 3
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Solve for "t":
t = u + 3
t = 6
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The number is 63
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Cheers,
Stan H.

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