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Math Help - solving a linear system by graphing

  1. #1
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    Exclamation solving a linear system by graphing

    I have no clue how to solve this
    Solve by graphing
    y=x+2
    y=4-x
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  2. #2
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    Quote Originally Posted by josh_c
    I have no clue how to solve this
    Solve by graphing
    y=x+2
    y=4-x
    Plot the line y=x+2, the on the same axes plot the line y=4-x.

    The solution is the point at which the two lines cross, as this is
    a point where the x and y values satisfy both equations, and hence
    is the required solution.

    RonL
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  3. #3
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    hi
    you've got to take a nice piece of paper and trace the lines (i dont know if its a proper expression) that are related to your two equation
    The solution should be the coordonates (x,y) of the intersection of those two lines so handle your rule corectly and dont shiver!
    You dont know how to trace a "line" related to an linear equation?
    no problemos:
    if you know to trace a line when youve got two points of the line i will tell you:
    ax+b=y is the general form of a linear equation
    so when you fix x you should know how to calculate y (if you dont give me your account number and your telephon number i will show you a trick)
    so you choose two x for which the point (x,y) (or (x,ax+b) if you preffer) are easy to plot on you piece of paper and there you go
    By
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  4. #4
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    Quote Originally Posted by CaptainBlack
    Plot the line y=x+2, the on the same axes plot the line y=4-x.

    The solution is the point at which the two lines cross, as this is
    a point where the x and y values satisfy both equations, and hence
    is the required solution.

    RonL
    i no how to plot this on the graph now im just not sure what do do after i plot it like how is the equation solved sorry i kinda sound dumb but i need help
    Last edited by josh_c; January 25th 2006 at 08:53 AM.
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  5. #5
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    Quote Originally Posted by josh_c
    i no how to plot this on the graph now im just not sure what do do after i plot it like how is the equation solved sorry i kinda sound dumb but i need help
    Look at the point at which the lines intersect. This point has an x-coordinate
    call this x_1 and a y-coordinate call this y_1.

    Then the solution is x=x_1 and y=y_1.

    To check that this is a solution plug these values back into the equations
    and you should find that they balance.

    RonL
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  6. #6
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    Quote Originally Posted by CaptainBlack
    Look at the point at which the lines intersect. This point has an x-coordinate
    call this x_1 and a y-coordinate call this y_1.

    Then the solution is x=x_1 and y=y_1.

    To check that this is a solution plug these values back into the equations
    and you should find that they balance.

    RonL
    The way i plotted it the lines do not intersect have i done this wrong or is there no solution
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  7. #7
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    re re hi
    so you want to understand what you are doing (take care this is no politicaly correct)

    When you plot a line (satisfying a linear equation) you plot on your paper
    all (a part indeed) the point which coordonates satisfy the equation
    why that? i have to think of it a few minuts to it (may be more)
    So you plot all the point satisfying equation 1
    and all the plot satisfying equation 2
    So if there is a point that satisfy the two equation it is the intersection of this two line you surely will understand why (you just have to say this point ...ect)
    if you know some litle basis of geometry you know that two unparralell line have only one intersection and that two parrallel line are the same line or have no intersection
    for the more tricky question wait a few minute
    or watch your boock
    re re By!
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  8. #8
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    Quote Originally Posted by SkyWatcher
    re re hi
    so you want to understand what you are doing (take care this is no politicaly correct)

    When you plot a line (satisfying a linear equation) you plot on your paper
    all (a part indeed) the point which coordonates satisfy the equation
    why that? i have to think of it a few minuts to it (may be more)
    So you plot all the point satisfying equation 1
    and all the plot satisfying equation 2
    So if there is a point that satisfy the two equation it is the intersection of this two line you surely will understand why (you just have to say this point ...ect)
    if you know some litle basis of geometry you know that two unparralell line have only one intersection and that two parrallel line are the same line or have no intersection
    for the more tricky question wait a few minute
    or watch your boock
    re re By!
    yea k i understand how to plot it now and all but i think i might have done it wrong becase they did not intersect so i dont no if its no solution or if i did it wrong
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  9. #9
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    Quote Originally Posted by josh_c
    yea k i understand how to plot it now and all but i think i might have done it wrong becase they did not intersect so i dont no if its no solution or if i did it wrong
    Try this:
    Attached Thumbnails Attached Thumbnails solving a linear system by graphing-plot.jpg  
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  10. #10
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    Quote Originally Posted by CaptainBlack
    Try this:
    so if i had y=x+6 and y=2-x the solution would be y=4 and x=2
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  11. #11
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    Quote Originally Posted by josh_c
    so if i had y=x+6 and y=2-x the solution would be y=4 and x=2
    No y=4, x=-2

    RonL
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  12. #12
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    Quote Originally Posted by CaptainBlack
    No y=4, x=-2

    RonL
    ok, i understand now how would i solve something like 2x-y+9=0, x+y-3+=0 like whats with the zero
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  13. #13
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    Quote Originally Posted by josh_c
    ok, i understand now how would i solve something like 2x-y+9=0, x+y-3+=0 like whats with the zero
    Rewrite them with the y on one side of the equals sign and every thing
    else on the other side.

    So 2x-y+9=0, can be rewritten:

    y=2x+9,

    and similarly x+y-3=0 can be rewritten:

    y=3-x

    RonL
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  14. #14
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    Quote Originally Posted by josh_c
    so if i had y=x+6 and y=2-x the solution would be y=4 and x=2
    You can easily see if you found a solution. Because by substituting them into your equations you get a match. Thus, what you said is wrong, it does not check.

    Here are the graphs for what you said,
    Attached Thumbnails Attached Thumbnails solving a linear system by graphing-picture2.gif  
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