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x2+6y=9⇒xy+122y=9xy+12=18y...........(i) x3+2y=4⇒xy+63y=4xy+6=12y...........(ii) Now, [(i)-(ii)] we have xy+12=18-xy+6=12y=6=6y=y=1 putting the value of 'y' in (ii), we have x×1+6=12×1⇒x=12-6x=6(x,y)=(6,1)
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