7w-3
------ +3<5
2 how can I find the solution set?

- anonymous

7w-3
------ +3<5
2 how can I find the solution set?

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- anonymous

\[\left|7w-3 \right|\]

- anonymous

The 7w-3 is an absoulte value

- anonymous

Well, start by moving the 3, to the other side and clearing the fractional part.

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## More answers

- anonymous

3+7w-3=5 right?

- anonymous

3+7w-3<5 I mean

- anonymous

No.
\[\frac{|7w-3|}{2} + 3 < 5\]
\[\implies \frac{|7w-3|}{2} < 5 -3\]
\[\implies |7w-3|< 2(5 + 3)\]
Right?

- anonymous

Ack!

- anonymous

Should be
\[|7w -3| < 2(5-3)\]

- anonymous

7w-3<22

- anonymous

Did you get 7w<22/3?

- anonymous

No.

- anonymous

2*2 = 4.

- anonymous

That means that 7w is less than 4 units away from 3.

- anonymous

The part of the equation where you have 7w-3<2(5+3)
=7w-3<10+6

- anonymous

I corrected that. It should be 2(5-3)

- anonymous

Not 2(5+3)

- anonymous

So that means
-4 < 7w - 3 < 4

- anonymous

Your right. And from there 7w<4-3
=7w<1
=w<1/7?

- anonymous

No.

- anonymous

\[|7w - 3| < 4 \]\[\implies -4 < 7w -3 < 4\]

- anonymous

Then you add 3 to each part of the relation.

- anonymous

\[-4 + 3 < 7w - 3 + 3 < 4 + 3\]

- anonymous

Wow thats alot

- anonymous

Think about it. If I tell you that | x | < 1, That means that x is somewhere between 1 and -1.

- anonymous

Right?

- anonymous

right

- anonymous

So if I tell you that |x + 3| < 4, that means that x+3 is somewhere between -4 and 4

- anonymous

So now we simplify and we have:
\[-1 < 7w < 7\]

- anonymous

ok, so I thought when we were at the 7w-3<4, we were almost finished. But we have to bring the 4 to the other side also and do it again.

- anonymous

well you can work the two relations at the same time.

- anonymous

Dividing the whole thing by 7 we get...?

- anonymous

1/7

- anonymous

-1/7 is part of it. what's the other part?

- anonymous

-1/7

- anonymous

Right.

- anonymous

7/7=1

- anonymous

-1/7

- anonymous

So w can be anything between -1/7 and 1 and the original relation \[\frac{|7w -3|}{2}+3 < 5\] Will be true.

- anonymous

Yep. (-1/7,1) is the solution set.

- anonymous

ok, great. I understand now. Thanks for your help.

- anonymous

Just remember when you take off the absolute values that you have 2 possible solutions.
\[|x| = a \implies x \in \{a,-a\}\]
\[|x| < a \implies -a < x < a \]
\[|x| \le a \implies -a \le x \le a \]
\[|x| \ge a \implies x \le -a \ or\ a\le x \]

- anonymous

etc.

- anonymous

Would the technique be the same if the absolute value contained a fraction?
-2\[-2\left| 7-v/8 \right|+3-11?\]
First step move the 8 over?

- anonymous

\[2\left| 7-v/8 \right|\]

- anonymous

\[-2\left| 7-v/8 \right|-3\]

- anonymous

hello?

- anonymous

Sorry, was cleaning

- anonymous

Yes, it'd be the same technique move everything that's not inside the absolute value to the other side, then evaluate the absolute value by splitting it up into parts.

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