If the equation 2x+4y=2y+4x is solved for y in terms of x, is

22y−2y+2x1−2x=0 Putting 2y=t, we get t2−t+2x1−2x=0, where t1=2y1 and t2=2y2t1t2=2x1−2x2y1+y2=2x1−2xy1+y2=x+log2⁡1−2x