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Math Help - Linear Inequality

  1. #1
    Junior Member Rheanna's Avatar
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    Linear Inequality

    Trying to remember how to do this.

    2x+5y _>-10

    Can I put whatever numbers I want in there? Like 0,1,2 to come up the line?
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  2. #2
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    Quote Originally Posted by Rheanna View Post
    Trying to remember how to do this.

    2x+5y _>-10

    Can I put whatever numbers I want in there? Like 0,1,2 to come up the line?
    Not exactly sure what you mean, but yes, you can basically put any number you want for one of x and y and solve for the other to plot the points on the line. You can simply draw the line though
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Rheanna View Post
    Trying to remember how to do this.

    2x+5y _>-10

    Can I put whatever numbers I want in there? Like 0,1,2 to come up the line?
    Hi Rheanna,

    You can set up a table of values for x, and solve for y. Noticing that your inequality is greater than or equal to, the line will be graphed as a solid line as opposed to a dashed or broken line.

    The easiest way to graph this is to first translate the inequality in the form of y = mx + b, using the appropriate inequality symbol.

    2x+5y \ge -10

    5y \ge -2x-10

    y \ge -\frac{2}{5}x-2

    You can now plot the y-intercept at (0, -2). Use the slope to find a second point (travel down 2 and right 5 or up 2 and left 5).

    The solution is a region either above or below the line. If you are not sure which side to shade, pick a test point and insert it into the original inequality and test for validity. If it works, shade that side. If it doesn't, shade the other side.
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  4. #4
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    Quote Originally Posted by Rheanna View Post
    Trying to remember how to do this.

    2x+5y _>-10
    Solve for y. Graph the line for the related equation. Then, since this is a "greater than" inequality, shade above the line.
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  5. #5
    Junior Member Rheanna's Avatar
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    Ok than, find the slope first, than do it, I see what you did. I did that for one of the questions but guess i looked too much into this one. Thanks
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