1. ## Inequality sum.

$x - 3 > x + 2$

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.

2. ## Re: Inequality sum.

Originally Posted by yorkey
$x - 3 > x + 2$
Is this really the problem: $x - 3 > x + 2~?$

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

3. ## Re: Inequality sum.

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

4. ## Re: Inequality sum.

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

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

5. ## Re: Inequality sum.

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

6. ## Re: Inequality sum.

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.

7. ## Re: Inequality sum.

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?

8. ## Re: Inequality sum.

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

9. ## Re: Inequality sum.

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.

10. ## Re: Inequality sum.

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.

11. ## Re: Inequality sum.

Alright, thank you very much!