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From the given information, we have,

$x+y=18andy+2xβ=31βor,x+y=18=>y=18βx....(i)Also,y+2xβ=31β=>3x=y+2=>3xβyβ2=0....(ii)$

Solving by substitution method,

Substituting (i) in (ii), we get,

$3xβyβ2=0=>3xβ18+xβ2=0=>4x=20=>x=5$

Putting $x=5$ in equation (i), we get,

$y=18βx=>y=18β5=>y=13$

Thus, the original fractyion is $yxβ=135β$

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