How do you solve #\frac { x - 1} { x - 2} - 3\geq 0#?

2 Answers
M7chael
Oct 6, 2017

Update: x could be any real number but #x!=2#

Explanation:

First off, #x!=2# because then you would be dividing by 0.

So,

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

Then add 3 on both sides,

#(x-1)/(x-2)>=3#

Multiply both sides by #(x-2)#

#color(blue)(x-1>=3(x-2))#

Distribute

#x-1>=3x-6#

Combine like terms

#5>=2x#

Divide by 2 on both sides

#2.5>=x#

So #2.5>=x# or #x<=2.5# but #x!=2#

However since you do not know if #x-2# is negative or positive, when you multiply both sides by #x-2# (#color(blue)(blue)#) it could switch the sign if that value is negative.

Therefore you could have #x-1<=3(x-2)# at that step which would lead to #x>=2.5#.

So really x could equal any number except for 2.

Thanks George C.!

George C.
Oct 6, 2017

#x in (2, 5/2]#, i.e. #2 < x <= 5/2#

Explanation:

Given:

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

First combine the rational expression with the constant #-3# to find:

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

That is:

#(-2x+5)/(x-2) >= 0#

In order that the left hand side be non-negative, we need one of the following:

  • #-2x+5 >= 0# and #x-2 > 0#. Hence #x in (-oo, 5/2] nn (2, oo) = (2, 5/2]#

  • #-2x+5 <= 0# and #x-2 < 0#. Hence #x in [5/2, oo) nn (-oo, 2) = O/#

graph{(-2x+5)/(x-2) [-10, 10, -5, 5]}

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