I am curious how to proceed with this problem:

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

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

Thanks in advance.

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- Dec 9th 2008, 09:38 AMZineferAbsolute Values
I am curious how to proceed with this problem:

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

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

Thanks in advance. - Dec 9th 2008, 09:49 AMmasters
Hello Zinefer,

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

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

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

How's that for a start? - Dec 9th 2008, 10:02 AMZinefer
$\displaystyle -\frac{3}{4}<\frac{27x}{28}<\frac{3}{4}$

Would that be the answer?

Again, thanks for your time. - Dec 9th 2008, 10:17 AMmasters
Not exactly, go bact to this step:

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

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

- Dec 9th 2008, 10:33 AMZinefer
Thanks alot. The answer is:

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

$\displaystyle (-\infty, -\frac{7}{5}) \cup (\frac{7}{10},\infty)$ - Dec 9th 2008, 11:52 AMmasters