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Math Help - Help with fractions and inqualities

  1. #1
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    Help with fractions and inqualities

    I am doing a review for an exam I have this week and I am stumped. Hoping someone here might have some insight:

    Express your solution in set-builder notation and interval notation:

    1/3(3x + 5) < 1/6(x + 4)


    I have no idea what to do. Do I multiply 1/3 and 1/6 throughout the parenthesis? I feel like Id rather just get rid of the fractions because they are so messy...

    Could someone please list a step by step process of how to solve these type of problems?
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  2. #2
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    Quote Originally Posted by gurrry View Post
    I am doing a review for an exam I have this week and I am stumped. Hoping someone here might have some insight:

    Express your solution in set-builder notation and interval notation:

    1/3(3x + 5) < 1/6(x + 4)


    I have no idea what to do. Do I multiply 1/3 and 1/6 throughout the parenthesis? I feel like Id rather just get rid of the fractions because they are so messy...

    Could someone please list a step by step process of how to solve these type of problems?
    Your expression is ambiguous.

    Is it

    \displaystyle\frac{1}{3(3x+5)}<\frac{1}{6(x+4)} \ \ \ \text{or}  \ \ \ \frac{3x+5}{3}<\frac{x+4}{6}

    What ever the expression is supposed to be, I would multiple both sides by 6 1st.
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  3. #3
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    the expression is:

    1/3 * (3x + 5) < 1/6 * (x + 4)


    * = multiplication


    the 1/3 and 1/6 stand alone next to the ( ) equations
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  4. #4
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    Quote Originally Posted by gurrry View Post
    the expression is:

    1/3 * (3x + 5) < 1/6 * (x + 4)


    * = multiplication
    \displaystyle 6\cdot\frac{3x+5}{3}<6\cdot\frac{x+4}{6}\Rightarro  w\cdots

    What do you have after you multiple by 6?
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  5. #5
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    You are really confusing me... why are you multiplying both sides by 6?

    I think you multiplied 1/3 and 1/6 through their respective equations.. but why are you now multiplying the 6 by both sides? To get rid of the fraction?


    If thats the case I guess youd have...

    6x + 10 < x + 4 ???
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  6. #6
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    Quote Originally Posted by gurrry View Post
    You are really confusing me... why are you multiplying both sides by 6?

    I think you multiplied 1/3 and 1/6 through their respective equations.. but why are you now multiplying the 6 by both sides? To get rid of the fraction?


    If thats the case I guess youd have...

    6x + 10 < x + 4 ???
    Let a,b,c\in\mathbb{Z}, \ \ \ b\neq 0

    \displaystyle\frac{a}{b}\times c=\frac{ac}{b}

    Multiplying by 6 gets rid of the fraction. Now, all you need to do is get the variables on one side and the numbers on the other. What do you get when you do that?
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  7. #7
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    x < - 6/5 ??


    hahaha wow just checked the answers and its right... so let me just make sure this is right:


    step 1) distrubute the fractions through the parenthesis

    step 2) multiply each side by the LCD to get rid of the fraction

    step 3) get variable by itself and solve
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  8. #8
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    Quote Originally Posted by gurrry View Post
    x < - 6/5 ??
    Looks good.
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  9. #9
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    Quote Originally Posted by gurrry View Post
    x < - 6/5 ??


    hahaha wow just checked the answers and its right... so let me just make sure this is right:


    step 1) distrubute the fractions through the parenthesis
    Actually, that was NOT what you did.

    step 2) multiply each side by the LCD to get rid of the fraction
    You did this first.

    step 3) get variable by itself and solve
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