How do you simplify\[\dfrac{5x-15}{x-3}\]?

Hint: In this question, we have asked to simplify the given fractional term. In the question, first we will find the factors of the numerator and then we check if it is divisible by denominator. If it is divisible then we will get the simplest form of the given fractional term. After that we will check where the denominator becomes zero. If at any value, the denominator then the fractional term will be undefined.

Complete answer:
In this question, it is asked to simplify the fractional term \[\dfrac{5x-15}{x-3}\] into a simple term.
We can see that in the fractional term, in the numerator it is given (5x-15) and in the denominator it is given (x-3). In numerator and denominator, in both the cases both are in the form of linear equations. So, we can say that when numerator is divided by denominator then the value of quotient or the value we will get after dividing them, we will get the term as constant or we will get another polynomial. But, in most of the cases, we get a constant term.
Now, let us see in the numerator.
5x can be written as 5 multiplied by x or we can write \[5\times x=5x\]
15 can be written as 5 multiplied by 3 or we can write \[5\times 3=15\]
As we can see in the numerator that there is a common factor 5 in the equation.
5x-15=5(x-3)
Therefore, we can write
\[\dfrac{5x-15}{x-3}=\left( \dfrac{5\left( x-3 \right)}{x-3} \right)\]
\[\Rightarrow \dfrac{5x-15}{x-3}=\left( 5 \right)\]
Hence, the simplified form of \[\dfrac{5x-15}{x-3}\] is 5.
For, x-3=0
The fractional expression is undefined.
So, for x=3, the fraction is undefined.
Hence, the value of expression is 5 for all real numbers of x except 3.

Note: In this type of question, we should have to check if the denominator is zero or not. If the denominator is zero, then the expression is not the legal fraction because its overall value is undefined.
Method of checking denominator is zero or not:
If the fraction is given in the form of \[\dfrac{N}{D}\], then we will check if D=0 for the value or the values of x.