Respuesta :

[tex] \dfrac{49 - \frac{1}{r^2} }{7 - \frac{1}{r} } [/tex]

Rewrite the fraction as division:
[tex]= (49 - \dfrac{1}{r^2}) \div (7 - \dfrac{1}{r} )[/tex]

Make them into single fraction:
[tex]= \dfrac{49r^2 - 1}{r^2} \div \dfrac{7r - 1}{r} [/tex]

Change the divide fraction into multiplication fraction:
[tex]= \dfrac{49r^2 - 1}{r^2} \times \dfrac{r}{7r - 1} [/tex]

Factorise the difference of square a² - b² = (a + b) (a - b) :
[tex]= \dfrac{(7r+ 1)(7r - 1)}{r^2} \times \dfrac{r}{7r - 1} [/tex]

Cancel the common factors:
[tex]= \dfrac{(7r+ 1)}{r}[/tex]
jcherry99
Comment. The best way to do this is to let 1/r = a and 1/r^2 = a^2. Now rewrite the problem.

Solution
[tex] \frac{49 - a^2}{7 - a} \\ \frac{(7 - a)(7 + a)}{7 - a}\text{ Notice a cancellation can take place} [/tex]

There is a 7 - a in both numerator and denominator, so that cancel providing a does not equal 7. a cannot equal 7 because that will put a 7 in the denominator and that makes the whole fraction = something over 0 which is undefined.

Answer
So far what  we have is 
7 + a

But a = 1/r
So the answer can be 7 + 1/r

r can be anything but 0 [for this answer]
and 1/7 for the cancellation.

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