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Math Help - graph solution and interval notation

  1. #1
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    graph solution and interval notation

    5/2(a+2) <-6 and 3/4(a-2)<1

    I am stuck.. so far I did 5/2 x (a+2)/1 ... I believe that is correct so far but it seems wrong.

    Wouldn't it be... 5(a+2)=2 but I am unsure how this plugs back into 5/2(a+2) <-6

    Also confused on this one..... -1 < -2x + 4 less than or equal to 5, and this does split into 2 equations?

    is it right to look like -5<-2x?
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  2. #2
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    Quote Originally Posted by pippers View Post
    5/2(a+2) <-6 and 3/4(a-2)<1

    I am stuck.. so far I did 5/2 x (a+2)/1 ... I believe that is correct so far but it seems wrong.

    Wouldn't it be... 5(a+2)=2 but I am unsure how this plugs back into 5/2(a+2) <-6

    Also confused on this one..... -1 < -2x + 4 less than or equal to 5, and this does split into 2 equations?

    is it right to look like -5<-2x?
    Do you mean \frac{5}{2}(a+2) or \frac{5}{2(a+2)}?

    Please take a moment to check out http://www.mathhelpforum.com/math-he...-tutorial.html. It is a very easy way to clarify your questions with precision to receive fast and efficient help.
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  3. #3
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    this one
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  4. #4
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    Quote Originally Posted by pippers View Post
    5/2(a+2) <-6 and 3/4(a-2)<1

    I am stuck.. so far I did 5/2 x (a+2)/1 ... I believe that is correct so far but it seems wrong.

    Wouldn't it be... 5(a+2)=2 but I am unsure how this plugs back into 5/2(a+2) <-6

    Also confused on this one..... -1 < -2x + 4 less than or equal to 5, and this does split into 2 equations?

    is it right to look like -5<-2x?
    \frac{5}{2}(a+2)<-6 can indeed be simplified by multiplying both sides by 2, but where did the -6 go, and how did your < turn into =?

    I'd advise reading up on Solving Inequalities: An Overview, for a good brush up on some basic inequality solving techniques.

    I'll start you off on the second one here.

    -1 < -2x + 4

    -5 < -2x

    I don't know what you mean by split into 2 equations, it doesn't need to be! You are correct to here!

    -5 < -2x We can divide both sides by -2, getting \frac{-5}{-2}=\frac{5}{2}

    \frac {5}{2} > x Keep in mind that if you multiply or divide both sides by a negative number, the inequality changes sign, > to < and vice versa.

    Can you apply these techniques to your first question now?
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  5. #5
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    Also the tutorial I did not understand....

    That is why I am confused on the <-6 and 3/4 (a-2) <1

    I have no clue how to solve it I thought cross multiply by add a 1 under (a+2) making it a fraction but it seems wrong.

    Problem #2 was -1 <-2x+4 less than or equal to 5

    I got....

    -1<-2+4 less/equal 5 -2x+4 less/equal 5
    = 5/2 >x = x>-1/2

    [-1/2,5/2)
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  6. #6
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    Quote Originally Posted by pippers View Post
    Also the tutorial I did not understand....

    That is why I am confused on the <-6 and 3/4 (a-2) <1

    I have no clue how to solve it I thought cross multiply by add a 1 under (a+2) making it a fraction but it seems wrong.

    Problem #2 was -1 <-2x+4 less than or equal to 5

    I got....

    -1<-2+4 less/equal 5 -2x+4 less/equal 5
    = 5/2 >x = x>-1/2

    [-1/2,5/2)
    Again, please read the LaTeX tutorial, I mean this in the best possible way that it is taxing to try to read through that mess of text just to help. LaTeX is not hard to learn. For example;

    [tex]\frac{5}{2} (a+2)> -6[/tex] gives \frac{5}{2} (a+2)> -6

    Similarly [tex]-1 < -2x + 4 \leq 5[/tex] gives -1 < -2x + 4 \leq 5. Much easier for anyone to read your post and understand with accuracy the question you are asking rather than trying to decipher regular text. The attached latex pdf at http://www.mathhelpforum.com/math-he...-tutorial.html has all of the symbols you will use, and it is just a matter of typing in the code for the symbol to put it in the equation.

    \leq = \leq (less equal)
    \geq = \geq (greater equal)

    You can also click on any of the equations in picture form to see the code for it.

    For question 2, you are right in that case.

    For question 1, let's work through the simplification of the first inequality. But you can't expect to understand that entire inequality tutorial in the 8 minutes you took after I posted it. Read close.
    Keep in mind that  \frac{5}{2}(a+2)=(\frac{5}{2})(\frac{(a+2)}{1})=\f  rac{5(a+2)}{2} So if we multiply the whole thing by 2, the 2 cancels out.

    \frac{5}{2}(a+2) < -6

    5(a+2) < (-6)(2) Multiply both sides by 2 to get rid of the denominator.

    5(a+2) < -12

     a + 2 < \frac{-12}{5} Divide both sides by 5 to get rid of the 5 on the left side

    Can you take it from here?
    Last edited by mr fantastic; September 28th 2009 at 10:50 PM. Reason: Used noparse to show thet latex with complete tags.
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  7. #7
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    I am unsure what to do... divide the 2 from both sides?

    To be honest if I see all the steps at once I am than able to use it as an example to do the rest of the problems that relate to it for a better understanding.

    Why cannot you multiple the reciprocal of 5/2 which is 2/5 to -6?
    Last edited by pippers; September 26th 2009 at 12:58 PM.
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  8. #8
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    Quote Originally Posted by pippers View Post


    I am unsure what to do... divide the 2 from both sides?

    To be honest if I see all the steps at once I am than able to use it as an example to do the rest of the problems that relate to it for a better understanding.

    Why cannot you multiple the reciprocal of 5/2 which is 2/5 to -6?
    No, you would subtract 2 from both sides. If you recall, working with an inequality is much like working with an equality ( = ), except for the special rules such as reversing the sign when multiplying or dividing by a negative number, etc.

    And you absolutely can multiply by the reciprocal, which is equivalent to multiplying by 2 and then dividing by 5. I noticed you started by multiplying by 2 in your first attempt, which is why I showed it the long way. The same technique we just solved this inequality with can also be used to solve the next inequality in this question.

    Showing all the steps serves a great purpose of answering a question for an assignment, but if you can't rationale the steps here, even using the second question that I did show completely, how well will you be able to apply those concepts on a brand new question with different variations of numbers and variables such as in a test or in a later course?
    Last edited by Kasper; September 26th 2009 at 01:59 PM.
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  9. #9
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    Does a < -17? Just to make sure I am on the right track?
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  10. #10
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    Quote Originally Posted by pippers View Post
    Does a < -17? Just to make sure I am on the right track?
    Not quite,

    a + 2 < -\frac{12}{5}

    a < -\frac{12}{5} - 2

    <br />
a < -\frac{12}{5}-\frac{10}{5} Remember that 2 = \frac{2}{1} = \frac{10}{5} We can multiply top and bottom by 5 to bring the fraction to have a denominator of 5 so we can subtract it.
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  11. #11
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    So its... a < \frac{-22}{5}?
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  12. #12
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    Quote Originally Posted by pippers View Post
    So its... a < \frac{-22}{5}?
    Yes.
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