1. ## Absolute value Inequality

Hey all,
I have the following absolute value inequality;
$|3x| - 1 \le |2x-1|$
which i simplify to:
$|3x| - |2x-1| \le 1$
The critical values are, $x = 0$ and $x = 1/2$;
Not sure how to solve it from now on..

2. Originally Posted by Oiler
Hey all,
I have the following absolute value inequality;
$|3x| - 1 \le |2x-1|$
which i simplify to:
$|3x| - |2x-1| \le 1$
The critical values are, $x = 0$ and $x = 1/2$;
Not sure how to solve it from now on..
Take a visual approach: Draw the graphs of y = |3x| - 1 and y = |2x - 1| and use them as a guide ....

3. I have gone through that example and still cant seem to grasp the concept. Would I in this example consider the case when,
$x \le \frac{1}{2}$ (which would be false in this case)
and when,
$\frac{1}{2} > x$

Hmm, no idea..

4. Originally Posted by Oiler
I have gone through that example and still cant seem to grasp the concept. Would I in this example consider the case when,
$x \le \frac{1}{2}$ (which would be false in this case)
and when,
$\frac{1}{2} > x$

Hmm, no idea..
Did you try doing what was suggested in post #2?

5. The question suggests that we use the number line instead of a graph, but the grap has touches the x axes at 2/5 and -2. I cant see how I can obtain those two values from original question..

6. Originally Posted by Oiler
The question suggests that we use the number line instead of a graph, but the grap has touches the x axes at 2/5 and -2. I cant see how I can obtain those two values from original question..
You're meant to look at where the graphs intersect (the x-coordinates of these points are required) and then consider what the equation of the lines are at these intersection points.

7. Originally Posted by Oiler
Hey all,
I have the following absolute value inequality;
$|3x| - 1 \le |2x-1|$
which i simplify to:
$|3x| - |2x-1| \le 1$
We study the cases when $x\le0,\,0\le x<\dfrac12$ and $x>\dfrac12.$

For $x\le0$ the inequality becomes $-3x-(1-2x)\le1$ then $x\ge -2$ so $-2\le x\le0$ is a solution set.

For $0\le x<\dfrac12$ the inequality becomes $3x-(1-2x)\le1$ so $x\le\dfrac25$ then $0\le x\le\dfrac25$ is another solution set.

For $x>\dfrac12$ the inequality becomes $3x-(2x-1)\le1$ so $x\le0$ but since the assumption is $x>\dfrac12$ there's no solution here.

So the solution set for the original problem is $\left[ -2,0 \right]\cup \left[ 0,\dfrac{2}{5} \right]=\left[ -2,\dfrac{2}{5} \right].$