Question

Question #9c7dc

Answer 1

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

Explanation:

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`.