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Modulus - C3!

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

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(edited 10 years ago)
where does mod(2x-5) = x
Reply 2
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JodieW
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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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JodieW
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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 :smile: Do I need simultaneous equations?


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x=2x-5 and x=-2x+5
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.
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 :smile: 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.
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
Reply 11
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JodieW
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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.
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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Reply 15
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JodieW
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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.
Original post by JodieW
I got 5 and 5/3 as my values of x, hope this is correct.


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Yep :smile:. You can always check by subbing your values back into the equation l 2x-5 l = x
Original post by JodieW


You are welcome !

Your answers should be correct.

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