How do you solve and write the following in interval notation: #3/(x-1) - 4/x>=1#?

Let's rearrange the inequality

#3/(x-1)-4/x>=1#

#3/(x-1)-4/x-1>=0#

#(3x-4(x-1)-x(x-1))/(x(x-1))>=0#

#(3x-4x+4-x^2+x)/(x(x-1))>=0#

#(4-x^2)/(x(x-1))>=0#

#((2-x)(2+x))/(x(x-1))>=0#

Let #f(x)=((2-x)(2+x))/(x(x-1))#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-2##color(white)(aaaaaa)##0##color(white)(aaaa)##1##color(white)(aaaaa)##2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##2+x##color(white)(aaaaa)##-##color(white)(aa)##0##color(white)(aaa)##+##color(white)(aaa)##+##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaa)##-##color(white)(aa)####color(white)(aaaa)##-##color(white)(a)##||##color(white)(a)##+##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-1##color(white)(aaaaa)##-##color(white)(aa)####color(white)(aaaa)##-##color(white)(a)####color(white)(aa)##-##color(white)(a)##||##color(white)(a)##+##color(white)(aaa)##+#

#color(white)(aaaa)##2-x##color(white)(aaaaa)##+##color(white)(aa)####color(white)(aaaa)##+##color(white)(a)####color(white)(aa)##+##color(white)(a)####color(white)(aa)##+##color(white)(a)##0##color(white)(a)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aa)##0##color(white)(aaa)##+##color(white)(a)##||##color(white)(a)##-##color(white)(a)##||##color(white)(a)##+##color(white)(a)##0##color(white)(a)##-#

Therefore,

#f(x)>=0# when # x in [-2,0) uu (1, 2]#

graph{3/(x-1)-4/x-1 [-35.57, 37.5, -25.57, 10.98]}