How do you solve #(p+5)/(p^2+p)=1/(p^2+p)-(p-6)/(p+1)# and check for extraneous solutions?

2 Answers
Mar 23, 2017

Answer:

#p=1,4#

Explanation:

#(p+5)/[p(p+1)]=1/[p(p+1)]-(p-6)/(p+1)#
#(p+5)/[p(p+1)]=[1-p(p-6)]/[p(p+1)]# => #p!=0,-1#
#p+5=1-p^2+6p#
#p^2-5p+4=0#
#(p-1)(p-4)=0#
#p=1, 4# => check both solutions in the original equation to verify they both work and not extraneous solutions.
#p=1#:
#(1+5)/(1+1)=1/(1+1)-(1-6)/(1+1)#
#6/2=1/2-(-5)/2#
#3=(1+5)/2#
#3=3# => verified p=1 is a valid solution, do the same for 4.

Apr 10, 2017

Answer:

#color(blue)(p=1# or #color(blue)(p=4#

Explanation:

#(p+5)/(p^2+p)=1/(p^2+p)-(p-6)/(p+1)#

#:.(p+5)/(p(p+1))=1/(p(p+1))-(p-6)/(p+1)#

#:.(p+5=1-p(p-6))/(p(p+1))#

#:.(p+5)/(p(p+1))=(1-p(p-6))/(p(p+1))#

multiply L.H.S. and R.H.S. by #color(blue)( (p(p+1))/1#

#:. cancelcolor(blue)(p(p+1))^color(red)1/1 xx (p+5)/cancel(p(p+1))^color(red)1=(1-p(p-6))/cancel(p(p+1))^color(red)1 xx cancelcolor(blue)(p(p+1))^color(red)1/1 #

#:.p+5=1-p(p-6)#

#:.p+5=1-p^2+6p#

#:.p^2-5p+4=0#

#:.(p-1)(p-4)=0#

#:.color(blue)(p=1 or p=4#

substitute #color(blue)(p=1#

#:.(color(blue)1+5)/(color(blue)1^2+color(blue)1)=1/(color(blue)1^2+color(blue)1)-(color(blue)1-6)/(color(blue)1+1)#

#:.6/2=1/2-((-5)/2)#

#:.6/2=1/2+5/2#

#:.6/2=6/2#

substitute #color(blue)(p=4#

#:.(color(blue)4+5)/(color(blue)4^2+color(blue)4)=1/(color(blue)4^2+color(blue)4)-(color(blue)4-6)/(color(blue)4+1)#

#:.9/20=1/20-((-2)/5)#

#:.9/20=1/20+2/5# #(2/5 xx 4/4=8/20)#

#:.9/20=1/20+8/20#

#:.9/20=9/20#