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Prentice Hall Algebra 1-problems with inequalities

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    Prentice Hall Algebra 1-problems with inequalities

    My name is Maddie (Sheri's daughter) and I am having some trouble with Lesson 9-6 #31. I was able to graph the lines themselves, and realized that negative multiplied by negative is positive, and positive multiplied by positive is also positive in a graph. However, I am still confused at why it is shaded in the way the answer says, like two triangles. I can see the answer in the back of the book and it looks right, but I don't know how to get to it. If I check it with points on the graph, it makes sense when I plug in the x and y in the problem, but I think I'm missing something.

    #2
    Hi Maddie,
    I don't have time to look at this until Monday, but I will get back to you then. Someone may answer you before I can, so stay tuned!

    Cindy Davis
    Science and Math teacher at Highlands Latin School - Indianapolis
    ds-25 college graduate: autodidact, working to pay the bills
    ds-23 college graduate: 1st year med school
    dd-21 college senior: Nursing

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      #3
      Hi Maddie,
      This problem is pretty tricky, which is why it is in the "Challenge" section.

      You have one difficult part complete - graphing the lines (and you knew to make them dashed, because "equals" is not included).
      Note that the point of intersection is not included in the solution, but it is an important boundary point. It is the intersection of the two lines and is the "solution" to that system of equations. If you solve the system of equations: y = x - 3 and y = -x - 2, the solution set is (1/2, -5/2) We will use x = 1/2 as a convenient boundary value.

      Now we can determine why the solution is shaded above and below the dashed lines:
      In order for the product to be > zero, both solution regions have to be positive or both be negative.

      for x values > 1/2, x-y-3 is positive when y<x-3, so shade below the line y = x - 3, for all x > 1/2 (intersection point)
      for x values > 1/2, x+y+2 is positive when y>-x-2, so shade above the line y = -x - 2, for all x > 1/2

      for x values < 1/2, x-y-3 is negative when y>x-3, so shade above the line y = x - 3, for all x < 1/2 (intersection point)
      for x values < 1/2, x+y+2 is negative when y<-x-2, so shade below the line y = -x - 2, for all x < 1/2

      I hope this explanation helps. Remember when you are graphing solutions to inequalities, you graph the boundary lines and then test values on both sides to see which side includes the solution.
      Cindy Davis
      Science and Math teacher at Highlands Latin School - Indianapolis
      ds-25 college graduate: autodidact, working to pay the bills
      ds-23 college graduate: 1st year med school
      dd-21 college senior: Nursing

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