\[\dfrac{x^2+x-6}{x-2}=x+3\]

**In the view of part (a), is this equation correct:**

\[ lim_{x \rightarrow 2 } \space \dfrac{x^2 +x-6}{x-2} = lim_{x\rightarrow 2 }(x+3) \]

This problem aims to find the equation’s right **domain,** making it an **equivalent fraction.** The concepts required for this problem are related to **quadratic algebra** which includes **domain, range** interception, and **undefined functions.**

Now the **domain** of a function is the group of values that we are permitted to put into our **function,** where such group of values is represented by the **x** terms in a **function** such as **f(x)**. Whereas the **range** of a function is a group of values that the **function** accepts. When we **plug** in the **x** values in that **function,** it shoots out the **range** of that function in the form of a group of **values.**

## Expert Answer

We need to understand the value of **domain** because it helps to define a **relationship** with the **range** of the function.

**Part a:**

Let’s first **factorize** the **left-hand** side of the equation so it becomes easy to **solve** it:

\[=\dfrac{x^2 + x – 6}{x -2}\]

\[=\dfrac{x^2 + (3 – 2)x – 6}{x -2}\]

\[=\dfrac{x^2 + 3x – 2x – 6}{x -2}\]

\[=\dfrac{(x – 2)(x + 3)}{x -2}\]

So here we have a **common factor** $(x-2)$ which can be **canceled** out. Thus we have $(x+3)$ left on the **left-hand** side.

Note that we have **simplified** the **left-hand** side to be equal to the **right-hand** side of the equation. So if we plug $x = 2$ into the **expression** $x + 3$, we do not get an **undefined value,** which is okay. but doing the same for the expression $ \dfrac{x^2 + x-6}{x-2} $ gives us an **undefined value.**

This is because we would get a $0$ in the **denominator,** resulting in an **undefined value.**

Therefore we cannot say that:

\[\dfrac{x^2 + x – 6}{x -2}=x+3\]

Unless we make a **requirement** in the above **expression** that is:

\[x\neq 2\]

Our **expression** becomes:

\[\dfrac{x^2 + x – 6}{x -2}=x+3,\space x\neq 2\]

The above expression states that all **numerical values** are allowed as the **domain** of the function, with the **exclusion** of the value $2$ which explicitly result in an **undefined value.**

**Part b:**

Yes, the **expression** is correct since you can reach as **close** to $2$ as you desire and these **functions** will still be **equal.** At the **actual** value $x=2$, these $2$ functions become **unequal** as stated in part $a$.

## Numerical Result

The **domain** must be **mentioned** with the **expression,** otherwise it will result in an **undefined value.**

\[\dfrac{x^2 + x – 6}{x-2}=x+3,\space x\neq 2\]

## Example

What is wrong with this equation?

$\dfrac{x^2 + x – 42}{x-6}=x+7$

We understand that for a **fraction** to exist, the **denominator** must be a **positive number** and it should not be equal to $0$.

Since we don’t have **variables** on the **right-hand** denominator, $x+7$ is achievable for all values of $x$, whereas the **left-hand** side has a **denominator** of $x-6$. For $x-6$ to be a positive number:

\[x>6; x\neq 6\]

Thus, our **expression** becomes:

\[\dfrac{x^2 + x – 42}{x -6}=x + 7,\space x\neq 6\]