In the equation #(3y-8)/(5y-2)=(y-2)/(y+5)#, w have uin the denominator #5y-2# and#y+5#. Asdenominator cannot be zero, we cannot have #y=2/5# or #y=-5#.
Multiplying each side of #(3y-8)/(5y-2)=(y-2)/(y+5)# by #(5y-2)(y+5)#, we get
#(3y-8)(y+5)=(y-2)(5y-2)#
or #3y^2+7y-40=5y^2-12y+4#
or #2y^2-19y+44=0#
or #2y^2-8y-11y+44=0#
or #2y(y-4)-11(y-4)=0#
or #(2y-11)(y-4)=0#
i.e. either #2y-11=0# i.e. #y=11/2#
or #y-4=0# i.e. #y=4#
Observe that none of the solution is #y=2/5# or #y=-5#, which are notin domain. Hence our solution is #y=11/2# or #y=4#.
