If two vectors are in parallel, then the coefficients of #a# and #b# are in the same ratio. For example, #a+b# and #2a+2b# are in parallel, because both the #a# and #b# coefficient are doubled. Another example: #-3a+5b# and #6a - 10b# are in parallel, because both coefficients have been multiplied by #-2#.
So, we want to impose the following:
#(x-2)/(2x+1)=1/-1#
which you can read as "the ratio of the #a# coefficients must be equal to the ratio of the #b# coefficients", where of course the #a# coefficients are #x-2# and #2x+1#, and the #b# coefficients are #1# and #-1#.
Since the equation obviously rewrites as
#(x-2)/(2x+1)=-1#, we can multiply by #2x+1# to get
#x-2=-(2x+1)=-2x-1#
Isolating the #x# terms and the costants, we get
#3x=1#, and thus #x=-1/3#
Let's verify:
For #x=1/3#, the two vectors are
#(1/3-2)a+b# and #(2/3+1)a-b#, which equals to
#-5/3a+b# and #5/3a-b#. As you can see, the coefficients of #a# and #b# are changed in sign, which means that they have both been multiplied by #-1#, and the vectors are thus in parallel.
