What is (n^2-1)/(39n)*(13n)/(7n+7) ?

The solution to (n^2-1)/(39n)*(13n)/(7n+7) is (n-1)/(21)

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simplify (n^2-1)/(39n)*(13n)/(7n+7)

simplify

\frac{n ^{2} - 1}{39 n} \cdot \frac{13 n}{7 n + 7}

\frac{n - 1}{21}

Solution steps

\frac{n ^{2} - 1}{39 n} \cdot \frac{13 n}{7 n + 7}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{( n ^{2} - 1 ) \cdot 13 n}{39 n ( 7 n + 7 )}

Cancel the common factor:

n

= \frac{( n ^{2} - 1 ) \cdot 13}{39 ( 7 n + 7 )}

Factor the number:

39 = 13 \cdot 3

= \frac{( n ^{2} - 1 ) \cdot 13}{13 \cdot 3 ( 7 n + 7 )}

Cancel the common factor:

13

= \frac{n ^{2} - 1}{3 ( 7 n + 7 )}

Factor

n^{2} - 1 : (n + 1) (n - 1)

= \frac{( n + 1 ) ( n - 1 )}{3 ( 7 n + 7 )}

Factor

7 n + 7 : 7 (n + 1)

= \frac{( n + 1 ) ( n - 1 )}{3 \cdot 7 ( n + 1 )}

Cancel the common factor:

n + 1

= \frac{n - 1}{3 \cdot 7}

Multiply the numbers:

3 \cdot 7 = 21

= \frac{n - 1}{21}

18 + 6

(x + 3) (x + 1) (x - 2)

e^{0.9}

4 (1)

6 - 18

What is (n^2-1)/(39n)*(13n)/(7n+7) ?
The solution to (n^2-1)/(39n)*(13n)/(7n+7) is (n-1)/(21)