absolute value

**dhiab** · June 13th 2009, 05:32 AM

Solve in R :

**alexmahone** · June 13th 2009, 05:40 AM

$|2x+1|\geq|3-x|$

Squaring both sides,

$4x^2+4x+1\geq9-6x+x^2$

$(x+4)(3x-2)\geq0$

So, $x\geq-4$ and $x\geq\frac{2}{3}$
ie $x\geq\frac{2}{3}$

Or $x\leq4$ and $x\leq\frac{2}{3}$
ie $x\leq-4$

**stapel** · June 13th 2009, 06:48 AM

You have:

. . . . . $\left|2x\,+\,1\right|\,\leq\,\left|3\,-\,x\right|$

The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, $(-\infty,\, -1/2),\,$ $(-1/2,\,3),\,$ and $(3,\,\infty).$

Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have:

. . . . . $-(2x\, +\, 1)\, \leq\, 3\, -\, x$

. . . . . $-2x\, -\, 1\, \leq\, 3\, -\, x$

. . . . . $-1\, \leq\, 3\, +\, x$

. . . . . $-4\, \leq\, x$

Since x = -4 is less than -1/2, then this solution is valid: $-4\,\leq\,x\,\leq\,-1/2$ .

If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so:

. . . . . $2x\, +\, 1\, \leq\, 3\, -\, x$

. . . . . $3x\, \leq\, 2$

. . . . . $x\, \leq\, \frac{2}{3}$

Then the inequality works for $-1/2\, \leq\, x\, \leq\, 2/3$ .

Now check when x > 3:

. . . . . $2x\, +\, 1\, \leq\, x\, -\, 3$

. . . . . $x\, \leq\, -4$

But x cannot be both greater than three and less than minus four, so there is no solution on this interval.

. . . . . $\mbox{solution: }\, -4\, \leq\, x\, \leq\, \frac{2}{3}$

This can be confirmed by graphing, by the way. For further information, try here.

. . . . .

**alexmahone** · June 13th 2009, 07:13 AM

Originally Posted by stapel

You have:

. . . . . $\left|2x\,+\,1\right|\,\leq\,\left|3\,-\,x\right|$

The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, $(-\infty,\, -1/2),\,$ $(-1/2,\,3),\,$ and $(3,\,\infty).$

Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have:

. . . . . $-(2x\, +\, 1)\, \leq\, 3\, -\, x$

. . . . . $-2x\, -\, 1\, \leq\, 3\, -\, x$

. . . . . $-1\, \leq\, 3\, +\, x$

. . . . . $-4\, \leq\, x$

Since x = -4 is less than -1/2, then this solution is valid: $-4\,\leq\,x\,\leq\,-1/2$ .

If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so:

. . . . . $2x\, +\, 1\, \leq\, 3\, -\, x$

. . . . . $3x\, \leq\, 2$

. . . . . $x\, \leq\, \frac{2}{3}$

Then the inequality works for $-1/2\, \leq\, x\, \leq\, 2/3$ .

Now check when x > 3:

. . . . . $2x\, +\, 1\, \leq\, x\, -\, 3$

. . . . . $x\, \leq\, -4$

But x cannot be both greater than three and less than minus four, so there is no solution on this interval.

. . . . . $\mbox{solution: }\, -4\, \leq\, x\, \leq\, \frac{2}{3}$

This can be confirmed by graphing, by the way. For further information, try here.

. . . . .

Sorry to break your bubble but you've copied the question wrong

Counterexample: x=0 doesn't satisfy the OP's inequality.

**dhiab** · June 13th 2009, 07:29 AM

Anathor resolution :

**alexmahone** · June 13th 2009, 07:34 AM

Originally Posted by dhiab

Anathor resolution :

Your first post has the inequality reversed.

**dhiab** · June 13th 2009, 07:37 AM

Originally Posted by alexmahone

$|2x+1|\geq|3-x|$

Squaring both sides,

$4x^2+4x+1\geq9-6x+x^2$

$(x+4)(3x-2)\geq0$

So, $x\geq-4$ and $x\geq\frac{2}{3}$
ie $x\geq\frac{2}{3}$

Or $x\leq4$ and $x\leq\frac{2}{3}$
ie $x\leq-4$

Hello :

your resolution is wrong, a product AB is negative mean that A, B are of opposite sign

**dhiab** · June 13th 2009, 07:40 AM

hello i'm sorry alexmahone

**alexmahone** · June 13th 2009, 07:40 AM

Originally Posted by dhiab

Hello :

your resolution is wrong, a product AB is negative mean that A, B are of opposite sign

My solution is right. A product AB is positive means that A, B are of same sign.

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