simplify this problem

• : [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.