
How do you simplify and find the restrictions for \[\dfrac{20+40x}{20x}\]?
Answer
451.2k+ views
Hint: This type of problem is based on the concept of polynomial functions. First, we have to consider the whole function. Take 20 common from the numerator and denominator of the function. Cancel the common term 20 from the numerator and denominator. Then, we get a function \[\dfrac{1+2x}{x}\] which is the simplified form of the function. To find the restriction, set the denominator not equal to zero. Solve the denominator and find the value of x for which the denominator is equal to zero which are the restrictions of the function.
Complete step by step answer:
According to the question, we are asked to simplify and find the restrictions of \[\dfrac{20+40x}{20x}\].
We have been given the equation is \[\dfrac{20+40x}{20x}\]. -----(1)
We first have to first simplify the given equation.
We can express the function as
\[\dfrac{20+20\times 2x}{20x}\]
We find that 20 are common in the numerator of the function. Let us now take 20 common from the both the terms of numerator.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20\left( 1+2x \right)}{20x}\]
We find that 20 are common in both the numerator and denominator of the function. On cancelling 20, we get
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{1+2x}{x}\]
Now, look for any common terms to cancel out. We find that there are no more terms to cancel out or take common.
Therefore, the simplified form of the function is \[\dfrac{1+2x}{x}\].
We now have to find the restriction of the function (1).
We know that a function is not defined when the denominator is equal to 0. To find the restrictions of the function, we have to set the denominator not equal to 0.
\[\Rightarrow 20x\ne 0\]
To find the value of x, we have to divide the whole expression by 20.
\[\Rightarrow \dfrac{20x}{20}\ne \dfrac{0}{20}\]
We know that 0 divided by any term is 0.
Therefore, we get
\[\dfrac{20x}{20}\ne 0\]
On cancelling 20 from the numerator and denominator, we get
\[x\ne 0\]
Therefore, the restriction is \[x\ne 0\].
Hence, the simplified form of the function \[\dfrac{20+40x}{20x}\] is \[\dfrac{1+2x}{x}\] and the restriction is \[x\ne 0\].
Note:
We can check whether the restriction is correct or not.
Substitute x=0 in the function.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20+40\times 0}{20\times 0}\]
We know that 0 multiplied to any term is equal to 0.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20+0}{0}\]
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20}{0}\]
Here, we find that the function is not defined when x=0. Hence, the restriction of the function is \[x\ne 0\].
Complete step by step answer:
According to the question, we are asked to simplify and find the restrictions of \[\dfrac{20+40x}{20x}\].
We have been given the equation is \[\dfrac{20+40x}{20x}\]. -----(1)
We first have to first simplify the given equation.
We can express the function as
\[\dfrac{20+20\times 2x}{20x}\]
We find that 20 are common in the numerator of the function. Let us now take 20 common from the both the terms of numerator.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20\left( 1+2x \right)}{20x}\]
We find that 20 are common in both the numerator and denominator of the function. On cancelling 20, we get
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{1+2x}{x}\]
Now, look for any common terms to cancel out. We find that there are no more terms to cancel out or take common.
Therefore, the simplified form of the function is \[\dfrac{1+2x}{x}\].
We now have to find the restriction of the function (1).
We know that a function is not defined when the denominator is equal to 0. To find the restrictions of the function, we have to set the denominator not equal to 0.
\[\Rightarrow 20x\ne 0\]
To find the value of x, we have to divide the whole expression by 20.
\[\Rightarrow \dfrac{20x}{20}\ne \dfrac{0}{20}\]
We know that 0 divided by any term is 0.
Therefore, we get
\[\dfrac{20x}{20}\ne 0\]
On cancelling 20 from the numerator and denominator, we get
\[x\ne 0\]
Therefore, the restriction is \[x\ne 0\].
Hence, the simplified form of the function \[\dfrac{20+40x}{20x}\] is \[\dfrac{1+2x}{x}\] and the restriction is \[x\ne 0\].
Note:
We can check whether the restriction is correct or not.
Substitute x=0 in the function.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20+40\times 0}{20\times 0}\]
We know that 0 multiplied to any term is equal to 0.
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20+0}{0}\]
\[\Rightarrow \dfrac{20+40x}{20x}=\dfrac{20}{0}\]
Here, we find that the function is not defined when x=0. Hence, the restriction of the function is \[x\ne 0\].
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