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$xxβ3β$

According to the question,

$xxβ3β+x+2xβ3+2β=2029β$

$xxβ3β+x+2xβ1β=2029β$

$βx_{2}+2xx_{2}+2xβ3xβ6+x_{2}βxβ=2029β$

$βx_{2}+2x2x_{2}β2xβ6β=2029β$

$β40x_{2}β40xβ120=29x_{2}+58x$

$β11x_{2}β98xβ120=0$

We know that,

$x=2aβbΒ±b_{2}β4acββ$

$x=22β(β98)Β±(β98)_{2}β4Γ11Γ(β120)ββ$

$x=2298Β±14884ββ$

$x=2298Β±(122)β$

$x=10,11β12β$

When $x=10$

Original fraction is $xxβ3β=1010β3β=xxβ3β=107β$Β

When $x=11β12β$

Original fraction is $xxβ3β=11β12β11β12ββ3β$

$βxxβ3β=1245β$

Hence, original fraction is $107β,1245β$

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