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

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

Mathematics
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\[\left|7w-3 \right|\]
The 7w-3 is an absoulte value
Well, start by moving the 3, to the other side and clearing the fractional part.

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3+7w-3=5 right?
3+7w-3<5 I mean
No. \[\frac{|7w-3|}{2} + 3 < 5\] \[\implies \frac{|7w-3|}{2} < 5 -3\] \[\implies |7w-3|< 2(5 + 3)\] Right?
Ack!
Should be \[|7w -3| < 2(5-3)\]
7w-3<22
Did you get 7w<22/3?
No.
2*2 = 4.
That means that 7w is less than 4 units away from 3.
The part of the equation where you have 7w-3<2(5+3) =7w-3<10+6
I corrected that. It should be 2(5-3)
Not 2(5+3)
So that means -4 < 7w - 3 < 4
Your right. And from there 7w<4-3 =7w<1 =w<1/7?
No.
\[|7w - 3| < 4 \]\[\implies -4 < 7w -3 < 4\]
Then you add 3 to each part of the relation.
\[-4 + 3 < 7w - 3 + 3 < 4 + 3\]
Wow thats alot
Think about it. If I tell you that | x | < 1, That means that x is somewhere between 1 and -1.
Right?
right
So if I tell you that |x + 3| < 4, that means that x+3 is somewhere between -4 and 4
So now we simplify and we have: \[-1 < 7w < 7\]
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.
well you can work the two relations at the same time.
Dividing the whole thing by 7 we get...?
1/7
-1/7 is part of it. what's the other part?
-1/7
Right.
7/7=1
-1/7
So w can be anything between -1/7 and 1 and the original relation \[\frac{|7w -3|}{2}+3 < 5\] Will be true.
Yep. (-1/7,1) is the solution set.
ok, great. I understand now. Thanks for your help.
Just remember when you take off the absolute values that you have 2 possible solutions. \[|x| = a \implies x \in \{a,-a\}\] \[|x| < a \implies -a < x < a \] \[|x| \le a \implies -a \le x \le a \] \[|x| \ge a \implies x \le -a \ or\ a\le x \]
etc.
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?
\[2\left| 7-v/8 \right|\]
\[-2\left| 7-v/8 \right|-3\]
hello?
Sorry, was cleaning
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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