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Thread: (Int. Algebra) I need help with an inequality problem

  1. #1
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    (Int. Algebra) I need help with an inequality problem

    Hello All,

    So here's the problem. I got it right but I'm still not clear on flipping the sign. As I understand it, you flip the sign when you multiply or divide by a negative number. In the operation below, why is the sign "not" flipped (from -1/3)? That's the only I got wrong and admittedly, I still don't understand it.

    (Int. Algebra) I need help with an inequality problem-0903181319.jpg

    Thanks.

    ALex
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  2. #2
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    Re: (Int. Algebra) I need help with an inequality problem

    it's not flipped because you divide by 3, not negative 3.
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  3. #3
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    Re: (Int. Algebra) I need help with an inequality problem

    Ok... I understand your answer. I'm still a little confused because I thought that -1/3, 1/-3, and -(1/3) were all the same. Oh I think I see now... The entire inequality is divided by "3." The 3 doesn't suddenly change to "-3." The way I have it written makes it look like there are individual fractions. Right? Or am I still not getting it. Wow! you have to be really careful.
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  4. #4
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    Re: (Int. Algebra) I need help with an inequality problem

    Solve (3x + 1)/4 = 2, then solve then solve (3x + 1)/4 = -3 : get my drift?
    Thanks from alexcordero
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  5. #5
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    Re: (Int. Algebra) I need help with an inequality problem

    Quote Originally Posted by alexcordero View Post
    Ok... I understand your answer. I'm still a little confused because I thought that -1/3, 1/-3, and -(1/3) were all the same. Oh I think I see now... The entire inequality is divided by "3." The 3 doesn't suddenly change to "-3." The way I have it written makes it look like there are individual fractions. Right? Or am I still not getting it. Wow! you have to be really careful.
    let's just work through it

    $-3 \leq \dfrac{3x +1}{4} \leq 3$

    we first multiply everything by 4. inequality signs do not flip

    $-12 \leq 3x+1 \leq 12$

    we now subtract 1 from all sides.

    $-13 \leq 3x \leq 11$

    we now divide by 3, inequality signs do not flip

    $-\dfrac{13}{3} \leq x \leq \dfrac{11}{3}$
    Thanks from alexcordero
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  6. #6
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    Re: (Int. Algebra) I need help with an inequality problem

    Quote Originally Posted by romsek View Post
    let's just work through it

    $-3 \leq \dfrac{3x +1}{4} \leq 3$

    we first multiply everything by 4. inequality signs do not flip

    $-12 \leq 3x+1 \leq 12$

    we now subtract 1 from all sides.

    $-13 \leq 3x \leq 11$

    we now divide by 3, inequality signs do not flip

    $-\dfrac{13}{3} \leq x \leq \dfrac{11}{3}$
    Ok, thank you very much for the explanation. On another question: How did you get the inequality and fractions to display as such?

    Again, thanks for help.
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  7. #7
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    Re: (Int. Algebra) I need help with an inequality problem

    Quote Originally Posted by alexcordero View Post
    Ok, thank you very much for the explanation. On another question: How did you get the inequality and fractions to display as such?
    If you click the tab reply with quote you will see the LaTeX code.

    Here is some: -3 \leq \dfrac{3x +1}{4} \leq 3
    -12 \leq 3x+1 \leq 12

    You can warp the code \$ -12 \leq 3x+1 \leq 12\$ in dollar signs you get $ -12 \leq 3x+1 \leq 12$
    Thanks from topsquark and alexcordero
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  8. #8
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    Re: (Int. Algebra) I need help with an inequality problem

    Quote Originally Posted by Plato View Post
    If you click the tab reply with quote you will see the LaTeX code.
    Nice!! I'm going to find a LaTeX syntax cheat sheet. Thanks again.
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