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# Modulus - C3! watch

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

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2. where does mod(2x-5) = x
3. (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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4. (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.
5. (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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6. x=2x-5 and x=-2x+5
7. 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.
8. (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.
9. (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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10. (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.
11. (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
12. (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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13. (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.
14. (Original post by JodieW)
So the branches would be y=2x-5 and y=-2x+5?

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

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

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17. (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.
18. (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
19. You are welcome !

Your answers should be correct.

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