Question 215408
"The value of a two-digit number is twice as large as the sum of its digits.
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Equation #1:
10t+u = =2(t+u)
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If the digits were reversed, the resulting number would be 9 less than 5 times the original number.
Equation #2:
10u+t =5(10t+u)-9
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Find the original number."
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Rearrange the two equations:
#1: 8t-u = 0
#2: 49t-5u = 9
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Then:
u = 8t
Substitute for "u" to solve for "t":
49t-5(8t) = 9
9t = 9
t = 1
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Since u = 8t, u = 8
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The original number is 18
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Cheers,
Stan H.