If #p=(8ab)/(a+b)#, what is the numerical value of #(p+4a)/(p-4a) + (p+4b)/(p-4b)#?

1 Answer
May 25, 2018

#2#.

Explanation:

#p=((8ab)/(a+b))#

Substitute this inside,

#(((8ab)/(a+b))+4a)/(((8ab)/(a+b))-4a) + (((8ab)/(a+b))+4b)/(((8ab)/(a+b))-4b)#

Simplify,

#((8ab)/(a+b)+(4a^2+4ab)/(a+b))/((8ab)/(a+b)-(4a^2+4ab)/(a+b)) + ((8ab)/(a+b)+(4b^2+4ab)/(a+b))/((8ab)/(a+b)-(4b^2+4ab)/(a+b))#

Combine,

#((8ab+4a^2+4ab)/cancel(a+b))/((8ab-4a^2-4ab)/cancel(a+b)) + ((8ab+4b^2+4ab)/cancel(a+b))/((8ab-4b^2-4ab)/cancel(a+b))#

Cancel common denominator,

#(8ab+4a^2+4ab)/(8ab-4a^2-4ab) + (8ab+4b^2+4ab)/(8ab-4b^2-4ab)#

Simplify,

#(4a^2+12ab)/(-4a^2+4ab) + (4b^2+12ab)/(-4b^2+4ab)#

Factor,

#(cancel(4a)(a+3b))/(cancel(4a)(-a+b)) + (cancel(4b)(b+3a))/(cancel(4b)(-b+a)#

Simplify,

#(a+3b)/(b-a) + (b+3a)/(a-b)#

Common denominator,

#(a+3b)/(b-a) - (b+3a)/(b-a)#

Combine,

#(a+3b-b-3a)/(b-a)#

Simplify,

#(2b-2a)/(b-a)#

Factor,

#(2cancel((b-a)))/cancel(b-a)#

Hence,

#2# :D