First, note that 2x-32x−3 and x+1x+1 are in the denominators of rational terms, meaning that to avoid dividing by 00 we must have x!=3/2x≠32 and x!=-1x≠−1.
With that in mind, first we eliminate the denominators by multiplying by (2x-3)(x+1)(2x−3)(x+1)
x/(2x-3)+4/(x+1)=1x2x−3+4x+1=1
=>(2x-3)(x+1)(x/(2x-3)+4/(x+1))=(2x-3)(x+1)*1⇒(2x−3)(x+1)(x2x−3+4x+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}
