SOLUTION: Could you help me from here;
The value of a two digit number is twice as large as the sum of its digits. If the digits were reversed, the resulting number would be 9 less than 5 t
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Question 139476: Could you help me from here;
The value of a two digit number is twice as large as the sum of its digits. If the digits were reversed, the resulting number would be 9 less than 5 times the original number. Find the original number.
t = tens digit
u = units digit
10t+u = original number
10u+t = the number named when the digits were reversed
10u+t = 9-5(10t+u)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The value of a two digit number is twice as large as the sum of its digits. If the digits were reversed, the resulting number would be 9 less than 5 times the original number. Find the original number.
t = tens digit
u = units digit
10t+u = original number
10u+t = the number named when the digits were reversed
10u+t = 9-5(10t+u)
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EQUATION:
10t + u = 2(t + u)
10u + t = 5(10t+u)-9
-------------------------
Rearrange the equations:
8t -u = 0
49t - 5u = 9
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solve by substitution:
49t - 5(8t) = 9
9t = 9
t = 1
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Substitute to solve for "u":
u = 8t so u = 8
-------------------
Original Number 18
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Cheers,
Stan H.
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