For the given relation: #y=x+2#
we can note that #(x,y)=(1,3)# is one possible solution.
If #y# varied directly as #x# then multiplying the #x# value (in this case #1#) by any value (for example #7#) would require that the #y# value (in this case #3#) if multiplied by that same value would give a new solution pair.
For the given example with an initial solution pair: #(1,3)#
then #(1xx7,3xx7)=(7,21)# would also need to be a valid solution.
Bit, in this case, #(x,y)=(7,21)# is not a valid solution pair for #y=x+2#.
In general for #y# to vary directly as #x# it must be possible to express the relation as #y=color(green)mx# for some constant, #color(green)m#
