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$letthefractionbe(yxβ)x=2y+1and(yxβ)+(xyβ)=2.9(xyx_{2}+y_{2}β)=2.9x_{2}+y_{2}=2.9xy(2y+1)_{2}+y_{2}=2.9(2y+1)y4y_{2}+4y+1+y_{2}=2.9(2y_{2}+y)4y_{2}+4y+1+y_{2}=(1029β)(2y_{2}+y)10Γ(5y_{2}+4y+1)=58y_{2}+29y50y_{2}+40y+10=58y_{2}+29y8y_{2}β11yβ10=08y_{2}β16y+5yβ10=08y(yβ2)+5(yβ2)=0(yβ2)(8y+5)=0y=2andy=(8β5β)β΄2y+1=2Γ2+1=5β΄number(25β)$

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