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  • Nov 13, 2009, 05:09 PM
    SamLovesBrian14
    Linear Equations for Algebra 1
    I need help on these, I never understand things like this.

    how do I find the answer to linear equations such as..

    x+y=6
    y=x+1
    y=x-2
    x+2=y

    you help me and ill try to help you! :)
  • Nov 13, 2009, 05:12 PM
    ballengerb1

    I don't think this is possible "y=x+1, y=x-2
    "
  • Nov 14, 2009, 08:41 AM
    Unknown008

    Do... you have to graph these? :confused:
  • Nov 24, 2009, 03:18 PM
    rosanna-hope

    if its y = x then it goes through the center of the graph, when its y=x+1 you move the line or point up one place on the x axis

    and the same with the y=x-2 except instead of adding 1 you mius 2 from the center
  • Nov 26, 2009, 10:16 AM
    mathwiz3502

    Like everyone is saying, please be more specific, but here are the characteristics of the equations you posted in relation to x=y which is a 45 degree diagonal line going through the origin (0,0)
    1.x+y=6, 6 units above
    2.y=x+1, 1 unit above
    3.y=x-2, 2 units below
    4.x+2=y, 2 units above
    If this does not help, I recommend asking about the 6 parent functions to either us or your teacher.
  • Nov 26, 2009, 10:17 AM
    mathwiz3502
    x=y also goes from the bottom left corner to the top right corner
  • Nov 28, 2009, 06:32 AM
    mathwiz3502

    OH, my bad, x+y=6 is 6 units below
  • Nov 28, 2009, 06:38 AM
    mathwiz3502
    Sorry, x+y=6 is flipped over the y axis and up 6 units
  • Nov 28, 2009, 07:57 PM
    QLP

    Ok this is how I would think about it.
    If you take x+y=6 first:
    When x=0,y=6 so you can put a cross on the graph at (0,6)
    When y=0, x=6, so you can put another cross on the graph at (6,0)
    Now you have two points on the graph so you can draw a straight line through them.

    Repeat for all the other equations by making first x and then y equal to 0 to see what the other co-ordinate is and then you will have all the points you need to draw all the lines.

    If any part of that isn't clear I will try to clarify further.

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