# Absolute Values

• December 9th 2008, 09:38 AM
Zinefer
Absolute Values
I am curious how to proceed with this problem:

$|(5/7)x+(1/4)|<3/4$

I know I must remove the Absolute value, but how?

• December 9th 2008, 09:49 AM
masters
Quote:

Originally Posted by Zinefer
I am curious how to proceed with this problem:

$|(5/7)x+(1/4)|<3/4$

I know I must remove the Absolute value, but how?

Hello Zinefer,

$\left \vert \frac{5x}{7}+\frac{1}{4} \right \vert \ < \frac{3}{4}$

Remember this: $\vert a \vert < b$ means $-b < a < b$. So,

$-\frac{3}{4}<\frac{5x}{7}+\frac{1}{4}<\frac{3}{4}$

How's that for a start?
• December 9th 2008, 10:02 AM
Zinefer
$-\frac{3}{4}<\frac{27x}{28}<\frac{3}{4}$

• December 9th 2008, 10:17 AM
masters
Not exactly, go bact to this step:

$-\frac{3}{4}<\frac{5x}{7}+\frac{1}{4}<\frac{3}{4}$

You will want to solve for x.

Multiply each term by 28 to remove the fractions. Then subtract 7 from both sides and the middle.

Code:

 -21 < 20x + 7 < 21  -7      - 7  -7 --------------------- -28 < 20x    < 14
Now, divide each term by 20, reduce your fractions, and you will find the range of x.
• December 9th 2008, 10:33 AM
Zinefer

$-\frac{7}{5} < x < \frac{7}{10}$

$(-\infty, -\frac{7}{5}) \cup (\frac{7}{10},\infty)$
• December 9th 2008, 11:52 AM
masters
Quote:

Originally Posted by Zinefer
$-\frac{7}{5} < x < \frac{7}{10}$
$(-\infty, -\frac{7}{5}) \cup (\frac{7}{10},\infty)$
Actually x is the set of points in the interval $\left(-\frac{7}{5}, \frac{7}{10}\right)$.
The solution set could also be written $\left \{ x \vert -\frac{7}{5}< x < \frac{7}{10}\right\}$