Question 121085

{{{(4w-1)/(3w+6)-(w-1)/(3)=(w-1)/(w+2)}}} Start with the given equation



{{{((3w+6)(3)(w+2))((4w-1)/(cross(3w+6))-(w-1)/(cross(3)))=((3w+6)(3)(w+2))((w-1)/(cross(w+2)))}}} Multiply both sides by the LCD {{{(3w+6)(3)(w+2)}}}. Doing this will eliminate every fraction.



{{{(3)(w+2)(4w-1)-(3w+6)(w+2)(w-1)=(3w+6)(3)(w-1)}}} Distribute and multiply. Notice every denominator has been canceled out.




{{{(3)(4w^2+7w-2)-(3w+6)(w^2+w-2)=3(3w^2+3w-6)}}} Foil




{{{(3)(4w^2+7w-2)-(3w^3+9w^2-12)=3(3w^2+3w-6)}}} Multiply {{{3w+6}}} and {{{w^2+w-2}}} to get {{{3w^3+9w^2-12}}}




{{{12w^2+21w-6-3w^3-9w^2+12=9w^2+9w-18}}} Distribute




{{{-3w^3-6w^2+12w+24=0}}} Get everything to the left side



{{{-3(w^2-4)(w+2)=0}}} Factor the left side



Now set each factor equal to zero:


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


Now solve for w for each factor:


{{{w=2}}} or  {{{w=-2}}} 



So our possible solutions are:


{{{w=2}}} or  {{{w=-2}}} 


However, if we plug in {{{w=-2}}} into the original equation, we'll get a zero denominator. So {{{w=-2}}} is <b>not</b> a possible solution




-------------------------------------


Answer:



So our solution is {{{w=2}}}