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.
color(white)("XXX")for 3y+2=2x
color(white)("XXX")3(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 (cxxa,cxxb) is also a solution (for any constant c).
color(white)("XXX")Noting that (4,3) is a solution to 3y+2=2x
color(white)("XXX")We can check using c=5
color(white)("XXX")to se if (x,y)=(5xx4,5xx3)=(20,15) is a solution.
color(white)("XXX")3(15)+2 = 47 != 40 = 2(20)
color(white)("XXXXXX")So, 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:
color(white)("XXX")y=m*x for some constant m
color(white)("XXX")3y+2=2x can be transformed into
color(white)("XXXXXX")y= 2/3x-2/3
color(white)("XXX")but there is no way to dispose of the (-2/3) to get it into the above form.
