math calculus
posted by Shreya on .
7/(x2)(x+5)8/(x+5)(x3)
7(x3)8(x2)/(x2)(x+5)(x3)
7x218x+16/(x2)(x+5)(x3)
7x8x give me 1x and 21+16 give me 5. for numerator but it have to be 1/(x2)(x3)
i not know how to do this please explain process my way not different way.

You have to use brackets to show that both
x2 and x+5 are factors of the bottom, so ...
7/((x2)(x+5))  8/((x+5)(x3)
so the lowest common denominator is (x2)(x+5)(x3)
we get
[ 7(x3)  8(x2) ] /[(x2)(x+5)(x3)]
= [7x  21  8x + 16]/[(x2)(x+5)(x3)]
= (x  5)/[(x2)(x+5)(x3)]
= 1(x+5)/[(x2)(x+5)(x3)]
= 1/[(x2)(x3)] 
= (x  5)/[(x2)(x+5)(x3)]
i understand how u do this part
= 1(x+5)/[(x2)(x+5)(x3)]
i not understand how you get numerator 
I simply factored out a 1

sorry but that the only thing i don't understand how do u factor that out? could you please explains slowly

suppose you had
2x + 4
then it would be 2(x+2)
suppose you had
2x  4
then it would be 2(x+2)
suppose you had x 5
then it would be 1(x+5)
in each case, expand my answer and see what you get. 
thank you very much reiny for taking time to explain :) u took all confusion away :)