How do you combine #-8/p+p/(p+5)#?

Question

1017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-15T21:59:03

#(p^2-8p-40)/(p^2+5p)#

Switching the order round the other way:

#color(green)([p/(p+5)color(red)(xx1)] -[8/pcolor(red)(xx1)] )#

#color(green)([p/(p+5)color(red)(xxp/p)] -[8/pcolor(red)(xx(p+5)/(p+5))] )#

#color(green)([p^2/(p(p+5))] -[(8(p+5))/(p(p+5))] larr" denominators now the same"#

#color(green)((p^2-8(p+5))/(p(p+5))#

#color(green)((p^2-8p-40)/(p(p+5)) )#

#color(green)((p^2-8p-40)/(p^2+5p))#