# Thread: absolute value

1. ## absolute value

Solve in R :

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

Squaring both sides,

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

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

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

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

3. You have:

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

The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, $\displaystyle (-\infty,\, -1/2),\,$$\displaystyle (-1/2,\,3),\, and \displaystyle (3,\,\infty). Consider each interval separately. If x < -1/2, then 2x + 1 < 0 and 3 - x > 0, so you have: . . . . .\displaystyle -(2x\, +\, 1)\, \leq\, 3\, -\, x . . . . .\displaystyle -2x\, -\, 1\, \leq\, 3\, -\, x . . . . .\displaystyle -1\, \leq\, 3\, +\, x . . . . .\displaystyle -4\, \leq\, x Since x = -4 is less than -1/2, then this solution is valid: \displaystyle -4\,\leq\,x\,\leq\,-1/2. If -1/2 < x < 3, then 2x + 1 > 0 and 3 - x > 0, so: . . . . .\displaystyle 2x\, +\, 1\, \leq\, 3\, -\, x . . . . .\displaystyle 3x\, \leq\, 2 . . . . .\displaystyle x\, \leq\, \frac{2}{3} Then the inequality works for \displaystyle -1/2\, \leq\, x\, \leq\, 2/3. Now check when x > 3: . . . . .\displaystyle 2x\, +\, 1\, \leq\, x\, -\, 3 . . . . .\displaystyle x\, \leq\, -4 But x cannot be both greater than three and less than minus four, so there is no solution on this interval. . . . . .\displaystyle \mbox{solution: }\, -4\, \leq\, x\, \leq\, \frac{2}{3} This can be confirmed by graphing, by the way. For further information, try here. . . . . . 4. Originally Posted by stapel You have: . . . . .\displaystyle \left|2x\,+\,1\right|\,\leq\,\left|3\,-\,x\right| The insides of the absolute values change signs at x = -1/2 and at x = 3, so you have three intervals; namely, \displaystyle (-\infty,\, -1/2),\,$$\displaystyle (-1/2,\,3),\,$ and $\displaystyle (3,\,\infty).$

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

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

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

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

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

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

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

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

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

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

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

Now check when x > 3:

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

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

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

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

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.

5. Anathor resolution :

6. Originally Posted by dhiab
Anathor resolution :
Your first post has the inequality reversed.

7. Originally Posted by alexmahone
$\displaystyle |2x+1|\geq|3-x|$

Squaring both sides,

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

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

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

Or $\displaystyle x\leq4$ and $\displaystyle x\leq\frac{2}{3}$
ie $\displaystyle x\leq-4$
Hello : your resolution is wrong, a product AB is negative mean that A, B are of opposite sign

8. hello i'm sorry alexmahone

9. 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.