Given that, x+1y=32...............(i) and y+1x=3........(i) Now, (i)×y and (ii)×x and minus calculate xy+1-xy-1=3y2-3x =0=3y2-3x =3x=3y2 ∴x=y2................(iii) x to apply (iii) y+1y2=3 =y+2y=3 =y2-3y+2=0 =y2-2y-y+2=0 =y(y-2)-1(y-2)=0 =(y-2)(y-1)=0 now, y=2, x=22=1 y=1, x=12 ∴(x,y) = (12,1) and (1,2)
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