Rhys_M
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Can someone tell me what to do for part b below please?
Thank you

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mqb2766
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Id start off by sketching a few lines on there for different values of p. What are the critical values where the number of solutions change?

To go down the algebra route, split the modulus function up into the two half lines and see where (what values of p) each intersects with the line. However, a sketch of several lines is good to get some intuition.
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Rhys_M
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(Original post by mqb2766)
Id start off by sketching a few lines on there for different values of p. What are the critical values where the number of solutions change?

To go down the algebra route, split the modulus function up into the two half lines and see where (what values of p) each intersects with the line. However, a sketch of several lines is good to get some intuition
So i've split it into half lines and equated them to px+5 to get the equations

-x=px+1 and x+2=px

So how do I find which value of p means they intersect? Do I equate both equations, as this gives p=-1/3 which doesn't seem right.

And even once I've found p, where do I go from there to answer the question? Thanks
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mqb2766
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Each half line is valid over a domain, so x<=-2 and x>=-2. When you remove the modulus funcition, you have to put a constraint like this on the working. So find which values of p give valid intersection with each half line.

If you have sketched the critical cases,that should guide you, so upload if necessary?
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Rhys_M
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(Original post by mqb2766)
Each half line is valid over a domain, so x<=-2 and x>=-2. When you remove the modulus funcition, you have to put a constraint like this on the working. So find which values of p give valid intersection with each half line.

If you have sketched the critical cases,that should guide you, so upload if necessary?
Yea sorry I don't really get what you mean about sketching the critical cases - i was just trying to go down the algebraic route
Edit - i'm really confused, would you mind doing a worked solution just for bi so I understand what you're talking about ?
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mqb2766
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For the sketch, sketch the modulus function. Then draw the line(s) for different values of p. Even just a few small integer values -2,-1,0,1,2 would be enough to get you thinking about what values of p would cause the number of solutions to change between (0,1,2). Here it is in desmos for p=2
https://www.desmos.com/calculator/ozblff1pi2
which has a single solution. I could put slider on for p, but ... The value of p=1 is obviously interesting becaue ..., so is ...

Upload what did for the algebraic route?
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Rhys_M
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(Original post by mqb2766)
For the sketch, sketch the modulus function. Then draw the line(s) for different values of p. Even just a few small integer values -2,-1,0,1,2 would be enough to get you thinking about what values of p would cause the number of solutions to change between (0,1,2). Here it is in desmos for p=2
https://www.desmos.com/calculator/ozblff1pi2
which has a single solution. I could put slider on for p, but ... The value of p=1 is obviously interesting becaue ..., so is ...

Upload what did for the algebraic route?
I deleted my work for algebraic route cause i didn't get anywhere with it.
Can you explain the algebraic route please - i hate graphical methods.
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mqb2766
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It should take 5-10 minutes to redo the algebra for the intersection of a line with a couple of half lines. Even if it isn't correct. The above hint about solving each line intersection problem on the appropriate domain (x<=-2 and x>=-2) which is formed when you remove the modulus function is all you need. The constraint introduces the "half" part of the half line into the analysis. Pls upload what you've tried.

If you hate graphical methods / sketching, I guess you'll struggle more with setting up unusual problems such as this. However, Im not going to force it on you. Note here, Im not proposing you solve the problem graphically, rather to make sure you interpret / understand the original problem and make sure the algebraic problem is set up correctly. Its almost irrelevant that the solution is "trivial" (or the validation is easy) once you've done / understand the sketch.
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