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anonymous
 5 years ago
7w3
 +3<5
2 how can I find the solution set?
anonymous
 5 years ago
7w3  +3<5 2 how can I find the solution set?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left7w3 \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 7w3 is an absoulte value

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, start by moving the 3, to the other side and clearing the fractional part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No. \[\frac{7w3}{2} + 3 < 5\] \[\implies \frac{7w3}{2} < 5 3\] \[\implies 7w3< 2(5 + 3)\] Right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Should be \[7w 3 < 2(53)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That means that 7w is less than 4 units away from 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The part of the equation where you have 7w3<2(5+3) =7w3<10+6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I corrected that. It should be 2(53)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that means 4 < 7w  3 < 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your right. And from there 7w<43 =7w<1 =w<1/7?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[7w  3 < 4 \]\[\implies 4 < 7w 3 < 4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then you add 3 to each part of the relation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[4 + 3 < 7w  3 + 3 < 4 + 3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Think about it. If I tell you that  x  < 1, That means that x is somewhere between 1 and 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if I tell you that x + 3 < 4, that means that x+3 is somewhere between 4 and 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now we simplify and we have: \[1 < 7w < 7\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, so I thought when we were at the 7w3<4, we were almost finished. But we have to bring the 4 to the other side also and do it again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well you can work the two relations at the same time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Dividing the whole thing by 7 we get...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/7 is part of it. what's the other part?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/7<w<1 So would it be accurate that the solution set would become (1/7, 1)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So w can be anything between 1/7 and 1 and the original relation \[\frac{7w 3}{2}+3 < 5\] Will be true.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep. (1/7,1) is the solution set.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, great. I understand now. Thanks for your help.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just 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
 5 years ago
Best ResponseYou've already chosen the best response.0Would the technique be the same if the absolute value contained a fraction? 2\[2\left 7v/8 \right+311?\] First step move the 8 over?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\left 7v/8 \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\left 7v/8 \right3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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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