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

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

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

2. ## Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1 Originally Posted by alexcordero 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.

3. ## Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1 Originally Posted by alexcordero 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.

4. ## Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1 Originally Posted by alexcordero 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:
[LIST=1][*]Is my solution correct? No.
$\displaystyle -3 < -3x + 4 < 3$

$\displaystyle -7 < -3x < -1$

$\displaystyle \frac{-7}{-3} > \frac{-3}{-3}x > \frac{-1}{-3}$

$\displaystyle \frac{7}{3} > x > \frac{1}{3}$

$\displaystyle \frac{1}{3} < x < \frac{7}{3}$

- - - - - - - - - - - - - - - - - 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:

$\displaystyle \frac{1}{3} < x < \frac{7}{3}$

5. ## Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1 Originally Posted by alexcordero 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.
$\begin{array}{l} |3x - 4| < 3\\ - 3 < 3x - 4 < 3\\ 1 < 3x < 7\\ \frac{1}{3} < x < \frac{7}{3} \end{array}$

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

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.

7. ## Re: [Int. Algebra] More inequality assistance please. |-3x+4|-4<-1 Originally Posted by melis 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.