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Math Help - Graphing inequalities

  1. #1
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    Angry Graphing inequalities

    Can somebody please show me how to graph inequalities? I don't get it and my homework is way late so please HELP!

    x>4 - graph the solution for each set of inequalities

    3x[U]>2x-4 - solve and graph the solution set of each inequalities

    I just need for someone to show and explain how to do these and I can try to do the rest on my own. Any help will be greatly appreciated THANKS
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  2. #2
    MHF Contributor Quick's Avatar
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    Do you mean on a graph or number line?
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  3. #3
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    Quote Originally Posted by Quick
    Do you mean on a graph or number line?
    Yes
    ---
    In the first problem just, draw a circle on top of 4 (to show you exclude 4) and draw a lined arrow to the right.

    In the second one, solve for all x,
    3x>2x-4
    3x-2x>2x-2x-4
    x>-4
    That means a circle above -4 and draw an arrowed line to the right.
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  4. #4
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    Hello, foofergutierrez!

    You should be familiar with Solving Inequalities.

    Solve for the variable as you would with an equation except:
    . . when multiplying or dividing by a negative, reverse the inequality.


    Graph the solution for: .  x > 4

    This is the set of all numbers greater than 4.

    . . .  - - - - \circ\!\!=\!=\!=\!=\!=\!=
    . . . . . . . . . 4

    The circle (\circ) indicates that the 4 is not included.



    Graph the solution for: . 3x \,\geq \,2x - 4

    Solve for x:\;\;3x\:\geq \:2x - 4\quad\Rightarrow\quad x \:\geq \:-4

    This is the set of all numbers greater than or equal to -4.

    . . . - - \bullet\!\!=\!=\!=\!=\!=\!=
    . . . . . -4

    The solid dot (\bullet) indicates that the -4 is included.

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  5. #5
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    Red face THANKS but I need more help

    Ok I kind of understand what to do but here are some other practice equations I would like to see done before my attempt to work my problems so that I know that I am doing them right.

    7(x-3>5x-14

    2(x+28)<6(x)

    I also have this weird word problem that I cannot make any sense of.

    The cost for a long distance call is $.36 for the first minute and $.21 for each additional minute. Write an equality representing the number of minutes a person could talk without exceeding $3.00.


    Again, I thank all who can help me and if I could I would pay ou back somehow! You guys are awesome!!!
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  6. #6
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    Quote Originally Posted by foofergutierrez

    7(x-3)>5x-14
    Open parantheses,
    7x-21>5x-14
    Remove the x's
    2x-21>-14
    Add 21 to both sides,
    2x>7
    Divide by 2,
    x>3.5
    ---
    2(x+28)<6(x)]
    Open parantheses,
    2x+56<6x
    Remove the x's (subtract 2x),
    56<4x
    Divide by 4,
    14<x
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  7. #7
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    Wink Thank you Perfect Hacker

    Thanks a million! It makes perfect sense to me now, you are a lifesaver .
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  8. #8
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    Quote Originally Posted by foofergutierrez
    you are a lifesaver
    I know.
    From where I come from everyone call me "God".
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  9. #9
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    Hello again, foofergutierrez!

    The cost for a long distance call is $0.36 for the first minute
    and $0.21 for each additional minute.
    Write an equality representing the number of minutes a person could talk
    without exceeding $3.00.

    Let x be the number of minutes we can talk for $3.00 or less.

    We are charged 36\!\!\not\!c for the first minute.

    We are charged 21\!\!\not\!c each for the other x - 1 minutes.
    . . This costs us: . 21(x - 1) cents.

    Hence, the total charge is: . 36 + 21(x - 1) cents

    Since we do not want to exceed $3 (300\!\!\not\!c)
    . . we have: . 36 + 21(x - 1)\:\leq \:300

    Solve for x:\;\;36 + 21x - 21 \:\leq\:300

    . . . . . . . . . . . . . 21x + 15\:\leq\:300

    . . . . . . . . . . . . . . . . 21x \:\leq \:285

    . . . . . . . . . . . . . . . . . . x\:\leq \:13.5714...


    Therefore, we can talk up to 13 minutes without going over the $3 limit.

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