Question

How do you simplify and state the excluded values for $\dfrac{4x-4}{x-1}$?

Answer 1

Hint: We first explain the process of factorisation which is available for the numerator of $\dfrac{4x-4}{x-1}$. We take 4 as common from the $4x-4$. We then simplify the fraction by completing the division. We try to find the part which can make the expression invalid or undefined. We find the points on which the denominator part becomes equal to 0. Those points will be excluded from the domain of the expression $\dfrac{4x-4}{x-1}$.

Complete step by step answer:
The given equation needs to be simplified. We need to factorise the numerator of $\dfrac{4x-4}{x-1}$.
The only process that is available for this equation to factorise is to take a common constant out of the terms 4 and $4x$. The only possible term is the constant 4.
We are taking 4 as common from the numbers 4 and $4x$.
Therefore, the factorisation is $4x-4=4\left( x-1 \right)$.
Now we find $\dfrac{4x-4}{x-1}=\dfrac{4\left( x-1 \right)}{x-1}=4$.
The simplified value of $\dfrac{4x-4}{x-1}$ is 4.
We need to find the domain of the expression $\dfrac{4x-4}{x-1}$.
The condition being that the expression has to give a defined solution.
We know that the denominator of a fraction can never be 0.
So, the points which will be excluded from the domain of the expression $\dfrac{4x-4}{x-1}$ are the points which makes the denominator of the expression $\dfrac{4x-4}{x-1}$ equal to 0.
So, we need to find the value of $x$ for which $x-1=0$.
We add 1 to both sides of $x-1=0$.
$\begin{align}
  & x-1+1=0+1 \\
 & \Rightarrow x=1 \\
\end{align}$.
Therefore, this point $x=1$ will make the expression v invalid.
The domain of the expression will be \[\mathbb{R}- \left\{ 1 \right\}\].

Note: We can verify the result of the factorisation by taking an arbitrary value of x where $x=2$.
We put $x=2$ on the equation $\dfrac{4x-4}{x-1}$ and get $\dfrac{4\times 2-4}{2-1}=\dfrac{8-4}{1}=4$
Thus, verified the factorisation $\dfrac{4x-4}{x-1}=4$.