**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.

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