Inequality sum.

**yorkey** · October 13th 2012, 06:46 AM

$x - 3 > x + 2$

This could either become:
$= - x + 3 < x + 2$
$= 1 < 2x$
$= \tfrac{1}{2} < x$

OR:
$= x - 3 < -x -2$
$= 2x < 1$
$= x < \tfrac{1}{2}$

Why do these equations have different answers? What I did was change the side on one side of the equation, and solve for x.

**Plato** · October 13th 2012, 06:51 AM

Originally Posted by yorkey

$x - 3 > x + 2$

Is this really the problem: $x - 3 > x + 2~?$

If so, there is no solution because $-3\not>2~.$

**grillage** · October 13th 2012, 06:52 AM

If you change the sign, you must flip the inequality too.

**yorkey** · October 13th 2012, 07:04 AM

Sorry sorry, I did my stupid old thing again by not giving the whole sum. This is the whole sum:

Solve the inequality $\left | x - 3 \right | > \left | x + 2 \right |$

**yorkey** · October 13th 2012, 07:06 AM

The mark scheme gives the correct answer as being $x < 1/2$

**Plato** · October 13th 2012, 07:16 AM

Originally Posted by yorkey

Solve the inequality $\left | x - 3 \right | > \left | x + 2 \right |$
The mark scheme gives the correct answer as being $x < 1/2$

Square both, $x^2-6x+9>x^2+4x+4$ Solve that.

**yorkey** · October 13th 2012, 07:43 AM

Aha, I get the right answer then. So what was wrong with my method, and why didn't it work? Should I always use the 'square both sides' method when it comes to solving modulus functions like this then?

**Soroban** · October 13th 2012, 07:58 AM

Hello, yorkey!

$\text{Solve the inequality: }\:|x-3| \:>\: |x+2|$

You could graph the expressions.

$y_1 \:=\:|x-3|$ is a $\vee$ -shaped graph with its vertex at (3,0).

$y_2\:=\:|x+2|$ is a $\vee$ -shaped graph with its vertex at (-2,0).

When is $y_1$ above $y_2$ ?

Code:

                  |
              *   |       .
                * |     .
                  *   .
                  | o
                  .   .
    .           . |     .           .
      .       .   |       .       .
    y2  .   .     |         .   .  y1
  - - - - . - + - + - + - + - . - - - -
         -2  -1   |   1   2   3
                  |

Answer: . $x\,<\,\tfrac{1}{2}$

**yorkey** · October 13th 2012, 08:00 AM

Thank you! But I'm afraid I don't have time. It's my exams coming up, and this is but the shortest type of question that I can expect and it should take me only a minute or two.

**Plato** · October 13th 2012, 08:02 AM

Originally Posted by yorkey

Aha, I get the right answer then. So what was wrong with my method, and why didn't it work? Should I always use the 'square both sides' method when it comes to solving modulus functions like this then?

We never says always for mathematics problems.
It work here because $|a|>|b|\text{ if and only if }a^2>b^2$ .

Had it been $x-3>|2+x|$ then it does not work.

**yorkey** · October 13th 2012, 08:08 AM

Alright, thank you very much!

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