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.
