Question

|3x – 2| < 7

Answer 1

solve it

Answer 2

Hint:

|a - b| < c is equivalent to -c < a - b < c

Answer 3

where did the sqaure root come from?

Answer 4

I was just playing around =]

Answer 5

wio's solution looked fine to me? Why'd you delete it?

Answer 6

i didnt delete she did

Answer 7

@nyaa, you're funny

Answer 8

\(|a|=+\sqrt{a^2}\)

Answer 9

@nyaa actually it was incorrect\[
|f(x)| <a \implies -a<f(x)<a
\]

Answer 10

@Hero would you care to elaborate? Today isn't a good day, I would like to LEARN how to solve this kind of problem as quickly as possible

Answer 11

What he means is when you have: \[
|3x – 2| < 7
\]It becomes: \[
-7<3x – 2 < 7
\]

Answer 12

l = Sbsolute value i know that. <=less than and > equals greater than correct?

Answer 13

\[|3x – 2| < 7 \iff -7 < 3x - 2 < 7\]

Answer 14

so you just take the absolute value of the lonely number and place it on both sides? How tdo you know where to put each < & > sign?

Answer 15

\[|3x-2|=\sqrt{(3x-2)^2}<7\\(3x-2)^2<49\\-7<3x-2<7\]

Answer 16

That's a completion to wio's solution isn't it?

Answer 17

The general form was posted

Answer 18

@nyaa Yeah that's true, but it's a bit funky to do it that way. I try funky things sometimes when I shouldn't.

Answer 19

@nyaa, I know what you're getting at, but no one really teaches it that way. But good on you.

Answer 20

I must be a unique snowflake.

Answer 21

so if i had l 4x - 2 l <8 i would set it up like: l4x-2l = √(4x-2)^2<8?

Answer 22

^ @nyaa, see what you did?

Answer 23

Did I do it incorrectly?

Answer 24

It's not incorrect, but it doesn't get you much further...

Answer 25

But I just have trouble seeing why "l3x-2l <7" = -7> 3x - 2 >7......

Answer 26

mannnnn....dsfasdfbjkrgbjkrgj im so pissed x(

Answer 27

\[
[f(x)]^2<b^2 \implies -b<f(x) <b
\]Isn't any more intuitive than: \[
|f(x)| < b \implies -b<f(x) <b
\]

Answer 28

An intuitive explanation is that \(|3x-2|\) means the magnitude of the number symbolized by \(3x-2\), and if that's less than 7, then 3x-2 can only go 7 steps below 0 and seven steps above.

Answer 29

@azee53 Do you understand what \(| \; |\) does?

Answer 30

so the exponents don't exactly make a difference?

Answer 31

yes, l l is absolute value

Answer 32

First, consider \[
|x|= \begin{cases}
x&x\ge 0 \\
-x & x< 0
\end{cases}
\]Now condier:\[
|f(x)|= \begin{cases}
f(x) & f(x)\ge0 \\
-f(x) & f(x)< 0
\end{cases}
\]

Answer 33

I apologize if you feel like you have wasted your time. I just can't get myself to understand any of this. I will try to analyze my lesson once more but this isn't working for me.. I DO appreciate the assistance however. Again I'm sorry if I wasted your time.

Answer 34

\[
\begin{split}
|f(x)|<a
&\implies \begin{cases}
f(x) <a& f(x)\ge0 \\
-f(x) <a& f(x)< 0
\end{cases}
\\
&\implies \begin{cases}
f(x) <a& f(x)\ge0 \\
-a<f(x) & f(x)< 0
\end{cases} \\
&\implies
-a<f(x)<a
\end{split}
\]

Answer 35

I'm going to go through my lesson content. I appreciate the help @wio

Answer 36

Good luck.

Answer 37

Thank you! Again I am sorry.