Method 1 For Seeing This
A direct variation equation will always pass through the origin
i.e. (x,y)=(0,0) will always be a solution to the equation.
XXXfor 3y+2=2x
XXX3(0)+2≠2(0) so this condition is not satisfied.
Method 2 For Seeing This
For a direct variation equation, if (a,b) is a solution, then (c×a,c×b) is also a solution (for any constant c).
XXXNoting that (4,3) is a solution to 3y+2=2x
XXXWe can check using c=5
XXXto se if (x,y)=(5×4,5×3)=(20,15) is a solution.
XXX3(15)+2=47≠40=2(20)
XXXXXXSo, once again, we see that this is not a direct variation.
Method 3 For Seeing This
Any direct variation can be transformed into the form:
XXXy=m⋅x for some constant m
XXX3y+2=2x can be transformed into
XXXXXXy=23x−23
XXXbut there is no way to dispose of the (−23) to get it into the above form.