If 2x+y=6y and 3x−1=2y+1 then the value of (log⁡3−log⁡2)(x−y) is

Taking log, we have (x+y)log⁡2=y(log⁡2+log⁡3)or   xlog⁡2=ylog⁡3∴xlog⁡3=ylog⁡2=x−ylog⁡3−log⁡2=λ (say)---i also  (x−1)log⁡3=(y+1)log⁡2or   xlog⁡3−ylog⁡2=log⁡3+log⁡2Using Eq. (i), we get λ(log⁡3)2−(log⁡2)2=log⁡3+log⁡2 λ=1log⁡3−log⁡2∴ 1λ=log⁡3−log⁡2=log⁡32