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
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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.