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C3 Modulus Inequality

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Could anyone tell me where I've gone wrong? The book has a different answer. :smile:
Reply 1
Avatar for davros
davros
16
Original post by Fudge2
image.jpg

Could anyone tell me where I've gone wrong? The book has a different answer. :smile:



I'm just going out, but you need to look at the critical regions for BOTH moduli - you can't necessarily assume that one stays with a + in front, while the other side can have either + or - in front of it!

Look at where EITHER side needs a minus sign put in front and then sort out the appropriate regions to investigate :smile:
Reply 2
Avatar for vannson
vannson
12
On the right hand side of the right hand side you should've expanded modulus to get -2x - 2, try that
Reply 3
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vannson
12
A good tip is to always sketch graphs to see where the intersection points are, and whether your answers are roughly correct. It helps a lot, and you can avoid a lot of stupid mistakes :smile:
Reply 4
Avatar for Fudge2
Fudge2
OP
5
Original post by davros
I'm just going out, but you need to look at the critical regions for BOTH moduli - you can't necessarily assume that one stays with a + in front, while the other side can have either + or - in front of it!

Look at where EITHER side needs a minus sign put in front and then sort out the appropriate regions to investigate :smile:


Ok haha. I'm afraid my knowledge on this subject is kinda limited...critical regions? I'm not sure how to determine which one turns to a minus. Is there a better way of solving these?

Original post by vannson
On the right hand side of the right hand side you should've expanded modulus to get -2x - 2, try that


How do you know to do that?
Reply 5
Avatar for vannson
vannson
12
Original post by Fudge2
Ok haha. I'm afraid my knowledge on this subject is kinda limited...critical regions? I'm not sure how to determine which one turns to a minus. Is there a better way of solving these?



How do you know to do that?


Okay, that was more of a guess... Try sketching both of them. If my sketch is right, you should be solving -x + 2 = 2x + 2 and -x + 2 = -2x - 2... Does that work?
Reply 6
Avatar for Fudge2
Fudge2
OP
5
Original post by vannson
Okay, that was more of a guess... Try sketching both of them. If my sketch is right, you should be solving -x + 2 = 2x + 2 and -x + 2 = -2x - 2... Does that work?


It makes sense if I sketch them. Thanks :smile:
Reply 7
Avatar for davros
davros
16
Original post by Fudge2
Ok haha. I'm afraid my knowledge on this subject is kinda limited...critical regions? I'm not sure how to determine which one turns to a minus. Is there a better way of solving these?



How do you know to do that?


If you look at |x - 2| it behaves differently where x < 2 from where x > 2
Similarly, |x + 1| switches where x < -1 and x > -1

So you need to sort out what happens when x < -1, when -1 < x < 2 and when x > 2

Alternatively, the graphing method will work if you're happy with that,

But you need to know both methods in case a question doesn't give you a choice!

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