Question 253783
The given equation of the line is 2x-3y = 6
The equation of a line parallel to this line is
                             2x-3y+c = 0 ....(1)  where c is a constant.
Given, the y-intercept is 7.
That is the point (0,7) lies on the line (1).
Put x\0 and y=7 in (1)
We get,  2*0-3*7+c = 0
              -21+c=0
                  c=21
Therefore (1)=>, the equation of the required line is 2x-3y+21 = 0