what values of x that satisfy I5-2xI<=(less than equal to) 7

xoxo1415 · **Joined:** Wed, 17 Aug 2011 23:43:31 UTC **Posts:** 11

How do we solve inequalities such as the one mentioned above where there is absolute value?

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.

Shadow · **Posted:** Thu, 15 Sep 2011 05:38:41 UTC

mathematic wrote:

You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.

Square both sides.

$4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0$ , divide that annoying 4 out:

$x^2-5x-6\le 0$

factor:

$(x-6)(x+1)\le 0$

From here it's easy.

mathematic · **Joined:** Fri, 1 Jul 2011 01:17:26 UTC **Posts:** 322

Shadow wrote:

mathematic wrote:

You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.

Square both sides.

$4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0$ , divide that annoying 4 out:

$x^2-5x-6\le 0$

factor:

$(x-6)(x+1)\le 0$

From here it's easy.

It's a matter of opinion, but I think my way is easier.

Shadow · **Posted:** Thu, 15 Sep 2011 23:57:36 UTC

mathematic wrote:

Shadow wrote:

mathematic wrote:

You can rewrite it as -7 ≤ 5 - 2x ≤ 7.
-12 ≤ -2x ≤ 2, 12 ≥ 2x ≥ -2, 6 ≥ x ≥ -1.

Square both sides.

$4x^2-20x+25\le 49\iff 4x^2-20x-24\le 0$ , divide that annoying 4 out:

$x^2-5x-6\le 0$

factor:

$(x-6)(x+1)\le 0$

From here it's easy.

It's a matter of opinion, but I think my way is easier.

Cases are almost never easier, this way it is absolutely immediately clear what the intervals are. And in any case it's senseless to talk about it, both are available to the op to choose as he likes, this isn't a popularity of solutions contest.

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what values of x that satisfy I5-2xI<=(less than equal to) 7

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