Question #9c7dc

The first thing to acknowledge here si the fact that any solution interval that you find must not include the value #x = 1#.

That is the case because #x=1# would make the denominator equal to zero, which as you know cannot happen. So right from the start, you know that #x !=1#.

Now, in order to get rid of the denominator, multiply the right side of the inequality by #1 = (1-x)/(1-x)#. You can do that because we've already established that #x!=1#.

You will have

#(2x-6)/(1-x) < 2 * (1-x)/(1-x)#

This can be reduced to

#2x - 6 < 2 * (1-x)#

Distribute the #2# to get

#2x - 6 < 2 - 2x#

Add #2x# to both sides of the inequality

#2x - 6 + 2x < 2 - color(red)(cancel(color(black)(2x))) + color(red)(cancel(color(black)(2x)))#

#4x - 6 < 2#

Next, add #6# to both sides of the inequality

#4x - color(red)(cancel(color(black)(6))) + color(red)(cancel(color(black)(6))) < 2 + 6#

#4x < 8#

Divide both sides of the inequality by #4#

#(color(red)(cancel(color(black)(4)))x)/color(red)(cancel(color(black)(4))) < 8/4#

#x < 2#

Now, when you write the solution interval, do not forget to add the restriction! In interval notation, the solution will be

#x in (-oo, 2) "\" {1}#

That means that #x# can take any value that is smaller than #2# and not equal to #1#.