anonymous
  • anonymous
7w-3 ------ +3<5 2 how can I find the solution set?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\left|7w-3 \right|\]
anonymous
  • anonymous
The 7w-3 is an absoulte value
anonymous
  • anonymous
Well, start by moving the 3, to the other side and clearing the fractional part.

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