How do you solve #8/(x-1)<1#?

1 Answer
Dec 25, 2016

Answer:

The answer is #x in ] -oo,1 [ uu ] 9, +oo[#

Explanation:

We cannot do crossing over

So,

#8/(x-1)<1#, #=>#, #8/(x-1)-1<0#

#(8-(x-1))/(x-1)<0#

#(9-x)/(x-1)<0#

Let #f(x)=(9-x)/(x-1)#

We can do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##1##color(white)(aaaa)##9##color(white)(aaaa)##+oo#

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

#color(white)(aaaa)##9-x##color(white)(aaaaa)##+##color(white)(aaa)##+##color(white)(aaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaa)##+##color(white)(aaa)##-#

Therefore,

#f(x)<0#, when #x in ] -oo,1 [ uu ] 9, +oo[#