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Math Help - linear systems & graphs

  1. #1
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    linear systems & graphs

    i cant figure out this stuff an example would be
    x+y=2
    x-y=5
    and i am supposed to match it with a graph how would i graph the above equation?
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  2. #2
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    Rearrange so you get y=x+a then make x equal a number of your choice. Say, 0. Then you have half of the coordinates for a point on that linear function; a point on that line. The other half of the coordinate would be what y becomes when you replace x with the coordinate you chose.

    In this case (x, y) \rightarrow (0, y) and since y=x+a \rightarrow y=0+a then y=a. Your coordinates are in that case (0, a). Once you repeat this process with another value for x, you get two points on this line. To draw the graph, just draw a line through the two points (and onwards).

    Hope I could be of assistance.
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  3. #3
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    There are 2 equations to sketch, what do you need to find? The point where they intersect maybe?
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  4. #4
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    the point where they intersect
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  5. #5
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    The two functions intersect where y_1=y_2 basically.

    Rearrange the equations as x+a=x+b and solve what x is. That's the x-coordinate for the point of intersection. Use that in the original equations to find out what the y-coordinate is.
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  6. #6
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    i have
    x+2=x+b
    x+5=x+b
    how do i solve what x is?
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  7. #7
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    If

    x+y=2
    x-y=5

    then

    y=-x+2
    -y=5-x \rightarrow y=x-5


    which gives you x-5=-x+2
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  8. #8
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    Quote Originally Posted by clay ford View Post
    i have
    x+2=x+b
    x+5=x+b
    how do i solve what x is?

    There was a problem when you tried to change the equation from standard to slope-intercept form.

    You have two equations the first one, in slope-intercept form is:
    y(1) = -X + 2

    The second is:

    y(2) = X - 5

    Think of it this way, you're trying to find a number to replace with "x" in order to get the same "y" value from both equations.

    You can do this by combining the two equations.
    You assume that Y(1) = Y(2) because that's what you want. You could represent Y(1) and Y(2) with their expression.

    y(1) = Y(2)
    -X + 2 = X - 5

    Isolate the variable to solve for X. Then plug in the "x" value into one of the two original equation (in whatever form to get 'y'). If you did it correctly, it shouldn't matter whatever equations you plugged 'x' into. The Y value you get with the x value you pluged is the point at which the two lines intersect each other on a coordinate plane.
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