Question 1130249
{{{y + 2/3 = (2y-12)/(3y-9)}}}

{{{(3y + 2)/3 = (2y-12)/(3y-9)}}}............cross multiply

{{{(3y + 2)(3y-9) = 3(2y-12)}}}

{{{9y^2-27y+6y-18 = 6y-36}}}

{{{9y^2-27y+6y-18 - 6y+36=0}}}

{{{9y^2-27y+cross(6y)-18 -cross( 6y)+36=0}}}

{{{9y^2-27y+18=0}}}......simplify, both sides divide by {{{9}}}

{{{y^2-3y+2=0}}}..............factor completely

{{{y^2-y-2y+2=0}}}

{{{(y^2-y)-(2y-2)=0}}}

{{{y(y-1)-2(y-1)=0}}}

{{{(y - 2) (y - 1) = 0}}}

solutions:

if {{{(y - 2)  = 0}}}->{{{y=2}}}

if {{{ (y - 1) = 0}}} ->{{{y=1}}}




 check the solution:


{{{y + 2/3 = (2y-12)/(3y-9)}}}->{{{y=2}}}

{{{2 + 2/3 = (2*2-12)/(3*2-9)}}}

{{{6/3 + 2/3 = (4-12)/(6-9)}}}

{{{8/3 = -8/-3}}}

{{{8/3 = 8/3}}}-> true; so, {{{y=2}}} is a solution


{{{y + 2/3 = (2y-12)/(3y-9)}}}->{{{y=1}}}

{{{1 + 2/3 = (2*1-12)/(3*1-9)}}}

{{{3/3 + 2/3 = (2-12)/(3-9)}}}

{{{5/3 = -10/-6}}}

{{{5/3 = -5/-3}}}

{{{5/3 = 5/3}}}-> true; so, {{{y=1}}} is a solution


there are {{{no}}} extraneous values

your answer: {{{2}}}, {{{1}}}



{{{ graph( 600, 600, -10, 10, -10, 10, x + 2/3 , (2x-12)/(3x-9)) }}}