Answer :
To find the product of the fractions
[tex]\[ \frac{4n}{4n-4} \cdot \frac{n-1}{n+1}, \][/tex]
we need to simplify each part of the expression before multiplying them together.
Firstly, let's simplify [tex]\(\frac{4n}{4n-4}\)[/tex]:
1. Factor out a common term from the denominator:
[tex]\[ 4n - 4 = 4(n - 1). \][/tex]
2. Now the fraction becomes:
[tex]\[ \frac{4n}{4(n - 1)} = \frac{4n}{4(n - 1)}. \][/tex]
3. We can cancel out the common factor of 4:
[tex]\[ \frac{4n}{4(n - 1)} = \frac{n}{n - 1}. \][/tex]
So, the simplified form of [tex]\(\frac{4n}{4n-4}\)[/tex] is [tex]\(\frac{n}{n-1}\)[/tex].
Secondly, consider the fraction [tex]\(\frac{n-1}{n+1}\)[/tex]:
The fraction [tex]\(\frac{n-1}{n+1}\)[/tex] is already in its simplest form.
Now, we multiply the two simplified fractions together:
[tex]\[ \frac{n}{n-1} \cdot \frac{n-1}{n+1}. \][/tex]
Notice that [tex]\((n-1)\)[/tex] appears once in the numerator and once in the denominator, so they cancel each other out:
[tex]\[ \frac{n}{n-1} \cdot \frac{n-1}{n+1} = \frac{n \cancel{(n-1)}}{\cancel{(n-1)}(n+1)} = \frac{n}{n+1}. \][/tex]
Therefore, the product of the given fractions simplifies to:
[tex]\[ \boxed{\frac{n}{n+1}}. \][/tex]
[tex]\[ \frac{4n}{4n-4} \cdot \frac{n-1}{n+1}, \][/tex]
we need to simplify each part of the expression before multiplying them together.
Firstly, let's simplify [tex]\(\frac{4n}{4n-4}\)[/tex]:
1. Factor out a common term from the denominator:
[tex]\[ 4n - 4 = 4(n - 1). \][/tex]
2. Now the fraction becomes:
[tex]\[ \frac{4n}{4(n - 1)} = \frac{4n}{4(n - 1)}. \][/tex]
3. We can cancel out the common factor of 4:
[tex]\[ \frac{4n}{4(n - 1)} = \frac{n}{n - 1}. \][/tex]
So, the simplified form of [tex]\(\frac{4n}{4n-4}\)[/tex] is [tex]\(\frac{n}{n-1}\)[/tex].
Secondly, consider the fraction [tex]\(\frac{n-1}{n+1}\)[/tex]:
The fraction [tex]\(\frac{n-1}{n+1}\)[/tex] is already in its simplest form.
Now, we multiply the two simplified fractions together:
[tex]\[ \frac{n}{n-1} \cdot \frac{n-1}{n+1}. \][/tex]
Notice that [tex]\((n-1)\)[/tex] appears once in the numerator and once in the denominator, so they cancel each other out:
[tex]\[ \frac{n}{n-1} \cdot \frac{n-1}{n+1} = \frac{n \cancel{(n-1)}}{\cancel{(n-1)}(n+1)} = \frac{n}{n+1}. \][/tex]
Therefore, the product of the given fractions simplifies to:
[tex]\[ \boxed{\frac{n}{n+1}}. \][/tex]
