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!


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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