Solve #|x-1|-|2+x|=3# ?

1 Answer
Jun 26, 2016

Answer:

#x = -5#

Explanation:

#|x-1|-|2+x|=3#

Making #y = abs(2+x)+3# we have

#abs(x-1)=y-> {(x-1=y),(x-1=-y):}#

1) From #x-1=y-> x = y+1# so
#y = abs(2+y+1)+3=abs(y+3)+3# but

#y+3=abs(y+3)+6# and
#pm 1=1+6/(y+3)# for #y ne -3# so the only possibility is
#-2=6/(y+3)-> y = -6->x = -5#

2)From #x-1=-y->x=-(y+1) # so
#y = abs(y-1)+3# but
#y-1=abs(y-1)+2# then
#pm1=1+2/(y-1)# for #y ne 1# so the only possibility is
#-2=2/(y-1)->y=0->x=1#

Substituting in the main equation we certify the solution: #x = -5#