Question 1100461
Maybe {{{x-3>=0}}} , and then {{{abs(x-3)=x-3}}} ,
or maybe {{{x-3<0}}} and then {{{abs(x-3)=-(x-3)}}} .
 
If {{{x-3<0}}} ,
the equation simplifies to
{{{-(x-3)=2x+1}}} <--> {{{-x+3=2x+1}}} <--> {{{3-1=2x+x}}} <--> {{{2=3x}}} <--> {{{x=2/3}}} .
That is a solution of {{{-(x-3)=2x+1}}} ,
but is it a solution of the original equation?
Let's check.
If {{{x=2/3}}} , {{{x-3=2/3-3=-7/3}}}  and {{{abs(x-3)=7/3}}} .
At the same time, if {{{x=2/3}}} , {{{2x+1=2*(2/3)+1=4/3+1=7/3}}} .
Clearly, {{{x=2/3}}} makes {{{abs(x-3)=2x+1}}} true,
so {{{highlight(x=2/3)}}} is a solution of {{{abs(x-3)=2x+1}}} .
 
What about the case when {{{x-3>=0}}} ?
Will that lead to another solution?
If {{{x-3>=0}}} ,
the equation simplifies to
{{{x-3=2x+1}}} <--> {{{-3-1=2x-x}}} <--> {{{-4=x}}} .
{{{x=-4}}} is a solution to {{{x-3=2x+1}}} ,
but it is not a solution to the original equation.
It makes {{{x-3=-7}}} , {{{abs(x-3)=7}}} , and
{{{2x+1=2*(-4)+1=-8+1=-7}}} .
Clearly, if {{{x=-4}}} makes {{{abs(x-3)=7}}} , and {{{2x+1=-7}}} ,
it is not a solution to {{{abs(x-3)=2x+1}}} .
we call it an extraneous solution.