[Int. Algebra] More inequality assistance please. |-3x+4|-4<-1

Oct 2017
23
0
San Diego, Ca
Hello all, I think I almost have this but I need to fine tune something...

Here's the problem:
\(\displaystyle |-3x+4|-4<-1\)

Here's what I did:
1. \(\displaystyle -3<-3x+4<3\)

2. \(\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}\)

3. \(\displaystyle \frac{7}{3}>x>\frac{1}{3}\)

My questions:
  1. Is my solution correct?
  2. Does the solution need to be rearranged? Like this: \(\displaystyle \frac{1}{3}<x<\frac{7}{3}\)
  3. What does the solution set look like? Is this acceptable? \(\displaystyle [\frac{1}{3},\frac{7}{3}]\)

Thank you.
-ac
 

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
Hello all, I think I almost have this but I need to fine tune something...

Here's the problem:
\(\displaystyle |-3x+4|-4<-1\)

Here's what I did:
1. \(\displaystyle -3<-3x+4<3\)

2. \(\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}\)

3. \(\displaystyle \frac{7}{3}>x>\frac{1}{3}\)

My questions:
  1. Is my solution correct?
  2. Does the solution need to be rearranged? Like this: \(\displaystyle \frac{1}{3}<x<\frac{7}{3}\)
  3. What does the solution set look like? Is this acceptable? \(\displaystyle [\frac{1}{3},\frac{7}{3}]\)

Thank you.
-ac
1. Yes it is. Well done!
2. Yes it is conventional to write your solution as 1/3 < x < 7/3.
3. You need to use ( ) instead of [ ] to show that 1/3 and 7/3 are excluded.
 
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Plato

MHF Helper
Aug 2006
22,490
8,653
Hello all, I think I almost have this but I need to fine tune something...

Here's the problem:
\(\displaystyle |-3x+4|-4<-1\)

Here's what I did:
1. \(\displaystyle -3<-3x+4<3\)

2. \(\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}\)

3. \(\displaystyle \frac{7}{3}>x>\frac{1}{3}\)

My questions:
  1. Is my solution correct?
  2. Does the solution need to be rearranged? Like this: \(\displaystyle \frac{1}{3}<x<\frac{7}{3}\)
  3. What does the solution set look like? Is this acceptable? \(\displaystyle [\frac{1}{3},\frac{7}{3}]\)
See Here.
 
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Dec 2016
302
165
Earth
Here's the problem:
\(\displaystyle |-3x+4|-4<-1\)

Here's what I did:
1. \(\displaystyle -3<-3x+4<3\)

2. \(\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3} \ \ \ \ \) The directions of the inequality symbols must be reversed. You need extra steps.

3. \(\displaystyle \frac{7}{3}>x>\frac{1}{3}\)

My questions:
  1. Is my solution correct? No.


  1. \(\displaystyle -3 < -3x + 4 < 3\)

    \(\displaystyle -7 < -3x < -1 \)

    \(\displaystyle \frac{-7}{-3} > \frac{-3}{-3}x > \frac{-1}{-3} \)

    \(\displaystyle \frac{7}{3} > x > \frac{1}{3}\)

    \(\displaystyle \frac{1}{3} < x < \frac{7}{3}\)


    - - - - - - - - - - - - - - - - - OR - - - - - - - - - - - - - - - - -


    \(\displaystyle -3 < -3x + 4 < 3\)

    \(\displaystyle (-1)(-3) > (-1)(-3x + 4) > (-1)(3)\)

    \(\displaystyle 3 > 3x - 4 > -3 \)

    \(\displaystyle 7 > 3x > 1 \)

    \(\displaystyle 1 < 3x < 7 \)

    \(\displaystyle \frac{1}{3} < x < \frac{7}{3}\)



    - - - - - - - - - - - - - - - OR - - -- - - - - - - - - - - - - -


    \(\displaystyle -3 < -3x + 4 < 3\)

    \(\displaystyle -7 < -3x < -1\)

    splits into

    \(\displaystyle -7 < -3x \)

    Add 3x to each side, and add 7 to each side:

    \(\displaystyle 3x < 7 \)

    \(\displaystyle \dfrac{3x}{3} < \dfrac{7}{3}\)

    \(\displaystyle x < \dfrac{7}{3}\)

    and

    \(\displaystyle -3x < -1\)

    Add 3x to each side, and add 1 to each side:

    \(\displaystyle 1 < 3x\)

    \(\displaystyle \dfrac{1}{3} < \dfrac{3x}{3}\)

    \(\displaystyle \dfrac{1}{3} < x\)

    Putting these together, we have:

    \(\displaystyle \frac{1}{3} < x < \frac{7}{3}\)
 
Last edited:

Plato

MHF Helper
Aug 2006
22,490
8,653
Hello all, I think I almost have this but I need to fine tune something...
Here's the problem: \(\displaystyle |-3x+4|-4<-1\)
Here is an advanced easy solution.
Because it is true that $(\forall a) :|a|=|-a|$ we can rewrite the question.
$\begin{array}{l} |3x - 4| < 3\\ - 3 < 3x - 4 < 3\\ 1 < 3x < 7\\ \frac{1}{3} < x < \frac{7}{3} \end{array}$
 
Sep 2018
3
0
US
Yes your work and answer are correct and you rearranged it correctly. But the solution [1/3, 7/3] is not acceptable, your solution should be (1/3, 7/3) to show that 1/3 and 7/3 are not included in the solution set.
 
Dec 2016
302
165
Earth
Yes your work ------> No, that is false. See * below.

and answer are correct ------> No, not only is that false, you contradict yourself at the end of your post. See ** below.

and you rearranged it correctly. But the solution [1/3, 7/3] is not acceptable, your solution should be (1/3, 7/3) to show that 1/3 and 7/3 are not included in the solution set.
* This step written by the OP is wrong:

2. \(\displaystyle \frac{-7}{-3}<\frac{-3}{-3}x<\frac{-1}{-3}\)

The direction of the inequality symbols reverse at the same moment that sides are multiplied or divided by a negative number.
It is an incorrect step.

It has to be this instead:

\(\displaystyle \frac{-7}{-3}>\frac{-3}{-3}x >\frac{-1}{-3}\)


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


** You stated "... and answer are correct."

"[1/3, 7/3] is not acceptable, " as you stated, because it is the wrong answer. It gets marked wrong.
 
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