First, note that #2x-3# and #x+1# are in the denominators of rational terms, meaning that to avoid dividing by #0# we must have #x!=3/2# and #x!=-1#.

With that in mind, first we eliminate the denominators by multiplying by #(2x-3)(x+1)#

#x/(2x-3)+4/(x+1)=1#

#=>(2x-3)(x+1)(x/(2x-3)+4/(x+1))=(2x-3)(x+1)*1#

#(xcancel((2x-3))(x+1))/cancel(2x-3)+(4(2x-3)cancel((x+1)))/cancel(x+1)=(2x-3)(x+1)#

#=>x(x+1)+4(2x-3)=(2x-3)(x+1)#

#=>x^2+x+8x-12 = 2x^2 - 3x+2x-3#

#=>x^2+9x-12 = 2x^2-x-3#

#=>x^2-10x+9 = 0#

#=>(x-1)(x-9)=0#

#=>x=1# or #x=9#

As neither of these conflict with our initial conditions of #x!=3/2# and #x!=-1#, we have the final solution set #x in{1,9}#