# Question #9c7dc

Dec 14, 2016

$x \in \left(- \infty , 2\right) \text{\} \left\{1\right\}$

#### 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 $x \ne 1$.

Now, in order to get rid of the denominator, multiply the right side of the inequality by $1 = \frac{1 - x}{1 - x}$. You can do that because we've already established that $x \ne 1$.

You will have

$\frac{2 x - 6}{1 - x} < 2 \cdot \frac{1 - x}{1 - x}$

This can be reduced to

$2 x - 6 < 2 \cdot \left(1 - x\right)$

Distribute the $2$ to get

$2 x - 6 < 2 - 2 x$

Add $2 x$ to both sides of the inequality

$2 x - 6 + 2 x < 2 - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{2 x}}}$

$4 x - 6 < 2$

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

$4 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} < 2 + 6$

$4 x < 8$

Divide both sides of the inequality by $4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}} < \frac{8}{4}$

$x < 2$

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

$x \in \left(- \infty , 2\right) \text{\} \left\{1\right\}$

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