Question

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

Answer 1

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

Answer 2

The 7w-3 is an absoulte value

Answer 3

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

Answer 4

3+7w-3=5 right?

Answer 5

3+7w-3<5 I mean

Answer 6

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

Right?

Answer 7

Ack!

Answer 8

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

Answer 9

7w-3<22

Answer 10

Did you get 7w<22/3?

Answer 11

No.

Answer 12

2*2 = 4.

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

Not 2(5+3)

Answer 17

So that means

-4 < 7w - 3 < 4

Answer 18

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

Answer 19

No.

Answer 20

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

Answer 21

Then you add 3 to each part of the relation.

Answer 22

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

Answer 23

Wow thats alot

Answer 24

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

Answer 25

Right?

Answer 26

right

Answer 27

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

Answer 28

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

Answer 29

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.

Answer 30

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

Answer 31

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

Answer 32

1/7<w

Answer 33

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

Answer 34

-1/7<w<7/7

Answer 35

Right.

Answer 36

7/7=1

Answer 37

-1/7<w<1
So would it be accurate that the solution set would become (-1/7, 1)?

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

Just remember when you take off the absolute values that you have 2 possible solutions.

\[|x| = a \implies x \in \{a,-a\}\]
\[|x| < a \implies -a < x < a \]
\[|x| \le a \implies -a \le x \le a \]
\[|x| \ge a \implies x \le -a \ or\ a\le x \]

Answer 42

etc.

Answer 43

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?

Answer 44

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

Answer 45

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

Answer 46

hello?

Answer 47

Sorry, was cleaning

Answer 48

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.