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    Name:  ImageUploadedByStudent Room1381267684.962293.jpg
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Size:  190.6 KB I have no idea what c) is asking :/ Could anyone help me understand?
    EDIT: b), not c), sorry ^-^

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    where does mod(2x-5) = x
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    (Original post by ChildishHambino)
    where does mod(2x-5) = x
    Where does it equal the x axis? Isn't it just 5/2?


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    (Original post by JodieW)
    Where does it equal the x axis? Isn't it just 5/2?


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    It's not where it crosses the x-axis, you want where |2x-5| intersects the line y=x.
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    (Original post by brittanna)
    It's not where it crosses the x-axis, you want where |2x-5| intersects the line y=x.
    Oh! Thanks Do I need simultaneous equations?


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    x=2x-5 and x=-2x+5
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    Yep. Solve /2x-5/=x. A sketch graph is always helpful when dealing with mods. The question implies there will be two solutions, so y=x will probably meet both halves of the y=/2x-5/ graph.
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    (Original post by JodieW)
    Where does it equal the x axis? Isn't it just 5/2?


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    you would solve it in the same way that if that "x" was a "5" or and "8" or "a". It's basically saying when do y=mod(2x-5) and y=x intersect.
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    (Original post by ChildishHambino)
    you would solve it in the same way that if that "x" was a "5" or and "8" or "a". It's basically saying when do y=mod(2x-5) and y=x intersect.
    But because of the mod part, how do I solve it? Can I solve it as if it were 2x-5 on its own?


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    (Original post by JodieW)
    Oh! Thanks Do I need simultaneous equations?


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    Not really. I'd start off by drawing a sketch to see whether the line y=x crosses both branches. Then set the equation of each relevant branch equal to x and solve those equations.
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    (Original post by JodieW)
    But because of the mod part, how do I solve it? Can I solve it as if it were 2x-5 on its own?


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    Both x values in these equations x=2x-5 and x=-2x+5
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    (Original post by brittanna)
    Not really. I'd start off by drawing a sketch to see whether the line y=x crosses both branches. Then set the equation of each relevant branch equal to x and solve those equations.
    So the branches would be y=2x-5 and y=-2x+5?


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    (Original post by JodieW)
    But because of the mod part, how do I solve it? Can I solve it as if it were 2x-5 on its own?


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    Because it is a mod graph, there will be two branches: 2x-5 and -(2x-5). You need to set both of these equal to x (after sketching the graph to make sure it crosses both) and solve those equations.
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    (Original post by JodieW)
    So the branches would be y=2x-5 and y=-2x+5?


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    Yeah :yes:.
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    (Original post by PUremathematician)
    Both x values in these equations x=2x-5 and x=-2x+5
    Thanks!


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    I got 5 and 5/3 as my values of x, hope this is correct.


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    (Original post by JodieW)
    I got 5 and 5/3 as my values of x, hope this is correct.


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    Try substituting them back in to see if they work.
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    (Original post by JodieW)
    I got 5 and 5/3 as my values of x, hope this is correct.


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    Yep . You can always check by subbing your values back into the equation l 2x-5 l = x
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    You are welcome !

    Your answers should be correct.
 
 
 
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