How do you solve #abs(4-3x)= 4x + 6# and state the extraneous solutions?

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Answer 1 · 2015-04-22T16:18:54

#abs(4-3x) = 4x+6#

Consider the two possibilities for #(4-3x)#

Possibility 1:
#(4-3x)<0#
#rarr x>4/3#

#abs(4-3x) = 4x +6#
becomes
#3x-4 = 4x+6#
#x=-10#
but #x>4/3# for this condition to apply
so this solution is extraneous.

Possibility 2:
#(4-3x)>=0#
#rarr x<= 4/3#

#abs(4-3x)= 4x+6#
becomes
#4-3x = 4x+6#
#x = 2/7#

Since #x=2/7# satisfies the condition #x<= 4/3#
this is a valid solution