Question 121085
{{{(4w-1)/(3w+6)-(w-1)/3=(w-1)/(w+2)}}}


Step 1 is to put your variable on one side, so add {{{-((w-1)/(w+2))}}} to both sides of the equation


{{{(4w-1)/(3w+6)-(w-1)/3-(w-1)/(w+2)=0}}}


Now we need a common denominator.  Notice that {{{3w+6}}} factors to {{{3*(w+2)}}} and those two factors are the other two denominators.  Therefore, {{{3w+6}}} is our lowest common denominator.


We can leave the first term alone, but the second term has to be multiplied by {{{(w+2)/(w+2)}}} and the third term must be multiplied by {{{3/3}}}


{{{(4w-1)/(3w+6)-((w-1)(w+2))/(3w+6)-3(w-1)/(3w+6)=0}}}


Now is a good time to note that since {{{w+2}}} is a factor in the denominator, we need to exclude {{{-2}}} from the solution set because that would make a denominator or denominators in the original equation go to zero.


Multiply the binomial factors and distribute the 3


{{{(4w-1-w^2-w+2-3w+3)/(3w+6)=0}}}


Collect terms


{{{4-w^2=0}}}


Factor the difference of two squares:


{{{(2-w)(2+w)=0}}}


So, {{{2-w=0}}} or {{{2+w=0}}}, in other words, {{{w=2}}} or {{{w=-2}}}.  But remember, we have to exclude {{{-2}}} from the solution set because that value makes the original equation undefined.  In the process of applying the common denominator, we introduced the square of the variable and that process introduced an extraneous root.


{{{x=2}}} is the only root of the equation.