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If I have $|3x| < x + 4$, I break it into two cases and get that $x<2$ and $x>-1$ and the question is done (I think).

In my solutions I have another problem that asks me to solve $|2x+3| = 2-x$,

and it is solved in the same way, except once the two solutions are found, they are plugged back into the equation to see if L.S. = R.S., and only one is valid.

My question is why is this necessary for that second problem but not for the first one? I don't understand. Thanks in advance!

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The extraneous solution are common when you square both sides an equation or an inequality to get rid of square roots or absolute values. In either case it is a good practice to check the result to see if all the solutions or intervals of solutions are satisfactory.

In you first example $$|3x| < x+4$$, we consider two cases,

1) $x\ge 0$ where the inequality is equivalent to $3x<x+4$, or $0\le x <2$

2)$x<0$ where the inequality is equivalent to $-1<x<0$

As the result the solution interval is $$-1<x\le 2$$

Now we test the intervals by plugging a test value in our inequality to see if we did it right.

For example for $x=-2$ we get $6<2$ which is false.

For your second example $$|2x+3|=2-x$$

We consider two cases.

1) $2x+3<0$ where the equation is equivalent to x=-5

2) $2x+3\ge 0$ where the equation is equivalent to $x=-1/3$

We check our results and fortunately both both solutions work.

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