
How do you simplify the complex fraction \[\dfrac{cx+cn}{{{x}^{2}}-{{n}^{2}}}\]?
Answer
550.2k+ views
Hint: Take ‘c’ from the numerator as the common term from both the terms cx and cn. Write the remaining terms in the numerator inside the bracket. Now, leave the numerator and simplify the denominator using the formula: - \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]. Cancel the common terms from both the numerator and denominator to get the simplified form.
Complete step by step solution:
Here, we have been provided with the algebraic expression \[\dfrac{cx+cn}{{{x}^{2}}-{{n}^{2}}}\] and we are asked to simplify it. Let us use some basic algebraic identities to solve the above fraction.
Now, assuming the given expression as ‘E’ we have,
\[\Rightarrow E=\dfrac{cx+cn}{{{x}^{2}}-{{n}^{2}}}\]
First let us simplify the numerator of this expression. Clearly, we can see that the term ‘c’ is common in both the terms cx and cn, so taking ‘c’ common and writing the remaining term inside the bracket, we have,
\[\Rightarrow E=\dfrac{c\left( x+n \right)}{{{x}^{2}}-{{n}^{2}}}\]
Now, let us come to the denominator part of the fraction. As we can see that the expression in the denominator is of the form \[{{a}^{2}}-{{b}^{2}}\] which can be factored using the algebraic identity: \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], so we have,
\[\Rightarrow E=\dfrac{c\left( x+n \right)}{\left( x+n \right)\left( x-n \right)}\]
Cancelling the common factor which is \[\left( x+n \right)\] from both the numerator and the denominator, we get,
\[\Rightarrow E=\dfrac{c}{\left( x-n \right)}\]
Hence, the simplified form of the given expression is \[\dfrac{c}{\left( x-n \right)}\].
Note:
One may note that the given fraction is only defined when x and n are unequal. If x = n then the value of the denominator will become zero and the expression will become zero and the expression will become undefined. You must remember some basic algebraic identities like: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] etc. These are important formulas used everywhere in mathematics. Sometimes, if these formulas may not be applicable and we have to factor the terms of a quadratic expression then we use the middle term split method.
Complete step by step solution:
Here, we have been provided with the algebraic expression \[\dfrac{cx+cn}{{{x}^{2}}-{{n}^{2}}}\] and we are asked to simplify it. Let us use some basic algebraic identities to solve the above fraction.
Now, assuming the given expression as ‘E’ we have,
\[\Rightarrow E=\dfrac{cx+cn}{{{x}^{2}}-{{n}^{2}}}\]
First let us simplify the numerator of this expression. Clearly, we can see that the term ‘c’ is common in both the terms cx and cn, so taking ‘c’ common and writing the remaining term inside the bracket, we have,
\[\Rightarrow E=\dfrac{c\left( x+n \right)}{{{x}^{2}}-{{n}^{2}}}\]
Now, let us come to the denominator part of the fraction. As we can see that the expression in the denominator is of the form \[{{a}^{2}}-{{b}^{2}}\] which can be factored using the algebraic identity: \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], so we have,
\[\Rightarrow E=\dfrac{c\left( x+n \right)}{\left( x+n \right)\left( x-n \right)}\]
Cancelling the common factor which is \[\left( x+n \right)\] from both the numerator and the denominator, we get,
\[\Rightarrow E=\dfrac{c}{\left( x-n \right)}\]
Hence, the simplified form of the given expression is \[\dfrac{c}{\left( x-n \right)}\].
Note:
One may note that the given fraction is only defined when x and n are unequal. If x = n then the value of the denominator will become zero and the expression will become zero and the expression will become undefined. You must remember some basic algebraic identities like: - \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\], \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\] etc. These are important formulas used everywhere in mathematics. Sometimes, if these formulas may not be applicable and we have to factor the terms of a quadratic expression then we use the middle term split method.
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