You can put this solution on YOUR website! { { { 5/p^2 - 5/q^2 / 1/p + 1/q } } }
there is no need for the brackets ,,so remove them and we get ...
= 5/p^2 - 5/q^2 / 1/p + 1/q
thaking LCM a from frist two i.e. 5/p^2 - 5/q^2 and last two i.e. 1/p + 1/q
variables we get ...
=(5q^2 - 5p^2)/p^2q^2 / (q+p)/pq
on solving the equation we get..
=(5q^2 - 5p^2)/[pq *(q+p)]
taking commom from the numerator we get ...
= 5( q^2 - p^2)/ [pq * (q+p)]
now on applying formula a^2 - b^2 = (a+b)(a-b) we get...
= 5(q+p)(q-p)/ [pq * (q+p)]
[ (q+p) is common in both denominator and numerator so it will de eliminated ]and we get...
= 5(q-p) / pq
= (5q - 5p )/ pq
hope this will help you...
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You can put this solution on YOUR website! (5/p^2 - 5/q^2) / (1/p + 1/q)
=5[(q^2-p^2)/(p^2q^2)]divided by [(q+p)/(pq)]
Taking 5 out and finding the lcm in the part before the division symbol and finding the lcm in the part after the symbol
=5[(q+p)(q-p)/(p^2q^2)]divided by [(q+p)/(pq)]
= 5[(q+p)(q-p)/(p^2q^2)]X [(pq)/(q+p)]
When division symbol is replaced by mulitplication symbol the fraction that is after the symbol is reciprocated that is the original fraction after the symbol is replaced [1/the fraction]
= 5(q-p)/pq (Cancelling (p+q) and pq)
(