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Math Help - Please Help - Solving, Graphing, and Factoring

  1. #1
    wonderwall09
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    Please Help - Solving, Graphing, and Factoring

    I have 3 algebra problems that I have no idea how to work. It's the beginning of the school year and our teacher gave us a diagnostic worksheet to see what we know, but she still wants the answers to be correct. So, if you could tell me how to do each of these problems in the simplest way possible, I'd really appreciate it.

    1.) Solve 3y - 2 > 5y. Graph the solution.

    2.) Graph 2x + 4y < 1

    3.) Factor 2x^2 - 5x - 12
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by wonderwall09 View Post

    1.) Solve 3y - 2 > 5y. Graph the solution.
    Don't ya just hate all that nasty stuff that comes back to haunt you?

    3y - 2 > 5y

    -2y > 2

    \frac{-2y}{-2} < \frac{2}{-2} <-- Note the switch from > to < because we are multiplying/dividing both sides by a negative number!

    y < -1

    Your instructor will want one of these two methods for graphing. Both are fairly standard, though I think the second method might be used a bit more.

    Method 1:
    To sketch this, first sketch out a number line marking 0 and a few integers around it. Just above the line you want to draw a ) over the -1 to indicate that -1 is not part of the solution, then an arrow from the ) pointing toward the negative side of the number line.

    Method 2:
    To sketch this, first sketch out a number line marking 0 and a few integers around it. On the line, at the point y = -1 draw an open circle to indicate that -1 is not part of the solution, then an arrow from the open circle pointing in the negative side of the number line.

    Perhaps someone will be good enough to show you what the sketches look like.

    -Dan
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  3. #3
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    earboth's Avatar
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    Quote Originally Posted by wonderwall09 View Post
    I have 3 algebra problems that I have no idea how to work.
    ...
    2.) Graph 2x + 4y < 1

    3.) Factor 2x^2 - 5x - 12
    Hello,

    I hope this doesn't come too late ...

    to #2.):

    The inequality describes a half plane with a straight line as upper bound. The equation of the line is:

    y=-\frac{1}{2}x+\frac{1}{4}. The bound doesn't belong to the half plane. (see attachment)

    to #3.):

     2x^2-5x-12 =2\left(x^2-\frac{5}{2}x-6\right) =2\left(\left(x^2-\frac{5}{2}x+\frac{25}{16}\right) - \frac{25}{16}-6\right)
    = 2\left(\left(x-\frac{5}{4}\right)^2 - \frac{121}{16}\right) =2\left(\left(x-\frac{5}{4}\right)^2 - \left(\frac{11}{4}\right)^2\right) =2\left(x-\frac{5}{4} + \frac{11}{4}\right) \left(x-\frac{5}{4}- \frac{11}{4}\right)~=~ ~\boxed{(2x+3)(x-4)}
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