Question

How do you simplify and find the excluded values of \[\dfrac{6-y}{{{y}^{2}}-2y-24}\]?

Answer 1

Hint: This type of problem is based on the concept of polynomial functions. First, we have to consider the whole function. Convert the middle term of the denominator in such a way that we get common terms from both the first term and the last term. Here, we can write -2y as the sum of -6y and 4 y. Take y common from the first two terms of the denominator and 4 common from the last two terms of the denominator. Then, take y-6 common from the two terms. Cancel the common terms from the numerator and denominator and find the simplified form of the equation. To find the excluded values of the function, equate the denominator to zero. Solve the denominator and find the value of x for which the denominator is equal to zero which is the excluded value of the function.

Complete step by step solution:
According to the question, we are asked to simplify and find the excluded value of \[\dfrac{6-y}{{{y}^{2}}-2y-24}\].
We have been given the equation is \[\dfrac{6-y}{{{y}^{2}}-2y-24}\]. -----(1)
We first have to first simplify the given equation.
We have to split the middle term of the denominator in such a way that we get common terms from the first and last term of the denominator.
Here, we can express the middle term -2y as -2y=-6y+4y.
Therefore, we get
\[\dfrac{6-y}{{{y}^{2}}-2y-24}=\dfrac{6-y}{{{y}^{2}}-6y+4y-24}\]
Now, let us take y common from the first two terms of the denominator and 4 common from the last two terms of the denominator.
\[\Rightarrow \dfrac{6-y}{{{y}^{2}}-2y-24}=\dfrac{6-y}{y\left( y-6 \right)+4\left( y-6 \right)}\]
Now, we find that y-6 is common in both the terms of the denominator. On taking (y-6) common, we get
\[\Rightarrow \dfrac{6-y}{{{y}^{2}}-2y-24}=\dfrac{6-y}{\left( y-6 \right)\left( y+4 \right)}\]
Now, let is take -1 common from the numerator. We get
\[\dfrac{6-y}{{{y}^{2}}-2y-24}=\dfrac{-\left( y-6 \right)}{\left( y-6 \right)\left( y+4 \right)}\]
On cancelling (y-6) from the numerator and denominator of the function, we get
\[\dfrac{6-y}{{{y}^{2}}-2y-24}=\dfrac{-1}{y+4}\]
Therefore, the simplified form of the function is \[\dfrac{-1}{y+4}\].
We now have to find the excluded values of the function (1).
We know that a function is not defined when the denominator is equal to 0. To find the excluded values of the function, we have to equate the denominator equal to 0.
\[\Rightarrow {{y}^{2}}-2y-24=0\]
But we have found that \[{{y}^{2}}-2y-24=\left( y-6 \right)\left( y+4 \right)\].
Therefore, we get
\[\left( y-6 \right)\left( y+4 \right)=0\]
That is, either y-6=0 or y+4=0.
Let us first consider y-6=0
Let us add 6 on both the sides of the equation.
\[\Rightarrow y-6+6=0+6\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 6, we get
y=0+6
On further simplifications, we get
y=6
Now, let us first consider y+4=0
Let us subtract 4 from both the sides of the equation.
\[\Rightarrow y+4-4=0-4\]
We know that terms with the same magnitude and opposite signs cancel out. On cancelling 4, we get
y=0-4
On further simplifications, we get
y=-4
Therefore, the excluded values of y in the function are 6 and -4.

Hence, the simplified form of the function \[\dfrac{6-y}{{{y}^{2}}-2y-24}\] is \[\dfrac{-1}{y+4}\] and the excluded values of y are 6 and -4.

Note: Whenever we get such types of questions, we have to equate the denominator of the function equal to 0 to find the excluded values. We should not equate the denominator of the simplified form of the function. We might not get all the excluded values. In this case, when we substitute the simplified denominator equal to 0, we just get the excluded value as -4 which is an incomplete answer. Avoid calculation mistakes based on sign conventions.