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.
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?
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.
Now, divide each term by 20, reduce your fractions, and you will find the range of x.Code:-21 < 20x + 7 < 21 -7 - 7 -7 --------------------- -28 < 20x < 14