Question 173693
I would appreciate your help on the following problem: 
The two digits in the numerator of a fraction whose value is 4/7 are reversed in its denominator. 
I have 10t+u/10u+t=4/7
Rearrange:
7(10t+u) = 4(10u+t)
70t + 7u = 40u+4t
66t - 33u = 0
t = (1/2)u
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The reciprocal of the fraction is the value obtained when 16 is added to the original numerator and 5 is subtracted from the original denominator. 
(10u+t)/(10t+u) = [10t+u+16]/[10u+t-5]
Rearrange:
(10u+t)(10u+t-5) = (10t+u)(10t+u+16)
[(10u+t)^2 - 5(10u+t)] = [(10t+u)^2 + 16(10t+u)]

[100u^2 + 20tu + t^2 -50u - 5t] = [100t^2 + 20tu + u^2 + 160t + 16u]


[99u^2 - 99t^2 -66u - 165t] = 0

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Substitute t = (1/2)u into the last equation to solve for "u":

Comment: This is a messy problem.  I'll leave the substitution to you.
I get u = 2 so t = 1
Number: 12

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Cheers,
Stan H.