Algebra 1 (Reiny)

Will you please explain the steps I must take in order to graph these linear inequalities?

Graph each system of linear inequalities. Give two ordered pairs that are solutions and two that are not solutions.

1. y < 2x - 1
y > 2

2. x < 3
y > x -2

3. y => 3x
3x + y => 3

4. 2x - 4y <= 8
y > x -2

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asked by Victoria

  1. in general if you have
    y > mx + b, you would shade in the region above the line y = mx + b
    y < mx + b , you would shade in the region below the
    line y = mx + b

    if the inequality contains ≥ or ≤ , the would include the boundary line itself and make it solid
    if you have just > or < , use a dotted line to set the boundary.

    for y > c, shade in the region above y = c, a horizontal line
    for y < c, shade in below

    for x > c , shade in the region to the right of the vertical line x = c
    for x < c, shade in the region to the left of y = c

    so let's do 3. , you do the rest

    y ≥ 3x, the line y = 3x would be included.
    two points on y = 3x are (0,0) and (3,9)
    plot the points and draw a solid line through them
    Shade in the region above y = 3x
    3x+y ≥ 3
    y ≥ -3x + 3
    two points on that one are (0,3) and (3,-6)
    draw a solid line through those points, and extend the line to suit your graph
    Shade in the region above that line as well.

    Your solution is the region shaded by both.
    Now just pick any point in that common region for your first part answer, then two point anywhere outside the common region to show a non-solution.

    Since a point has to satisfy both inequations, it must be in the common region.
    A point could satisfy one inequation but not the other, so it would NOT be a solution.

    your graph should look like this
    http://www.wolframalpha.com/input/?i=plot++y+%3D+3x+%2C+y+%3D+-3x+%2B+3+
    with the triangular region at the top shaded in

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    posted by Reiny

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