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# Question Watch

1. For this question why do i have to look at the regions of the modulus function?
And how do i do it?
Do i consider each modulus graph and find the region in which it acts?

Thanks!
Attached Images

2. (Original post by Zenarthra)

For this question why do i have to look at the regions of the modulus function?
And how do i do it?
Do i consider each modulus graph and find the region in which it acts?

Thanks!
Well, how else would you do it?

The point is that if you look at different regions you can replace modulus signs by + or - as appropriate e.g. if you have something like |x + 1| then in the region x < - 1 this is -(x + 1) whereas elsewhere it is just x + 1.
3. (Original post by davros)
Well, how else would you do it?
prsom
4. (Original post by davros)
Well, how else would you do it?

The point is that if you look at different regions you can replace modulus signs by + or - as appropriate e.g. if you have something like |x + 1| then in the region x < - 1 this is -(x + 1) whereas elsewhere it is just x + 1.

Ahh ok, but i dont see why the modulus function is determined under these domains?

Thanks!
Attached Images

5. (Original post by Zenarthra)

Ahh ok, but i dont see why the modulus function is determined under these domains?

Thanks!
You seem to have misunderstood. You can't look at the terms separately.

I'll get you started. How can we rewrite the equation if we know that x>2?
6. (Original post by BabyMaths)
You seem to have misunderstood. You can't look at the terms separately.

I'll get you started. How can we rewrite the equation if we know that x>2?
Im guessing to combine it all onto 1 side but im not sure why.
Am i right?
7. (Original post by Zenarthra)
Im guessing to combine it all onto 1 side but im not sure why.
Am i right?
The point is that when we have

|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2

so the equation can be written as..?
8. (Original post by BabyMaths)
The point is that when we have

|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2

so the equation can be written as..?
Im sorry but how did you deduce that if x> or = 2 then
|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2 ?

Thanks!
9. If x>2 then x+1>3 and so |x+1|=x+1.
10. (Original post by Zenarthra)
Im sorry but how did you deduce that if x> or = 2 then
|x+1|=x+1

|x|=x

|x-1|=x-1

and

|x-2|=x-2 ?

Thanks!
When x>2, all those expressions have a value greater than or equal to 0, so are not reflectes in the x-axis. Therefore the modulus graph of y=x-2 when x>2 has equation y=x-2 and not y=-(x-2) or*2-x. This is true for all the expressions, so once you've stated your range considerations, you can effectively ignore the mod signs and solve.
11. And I thought this was going to lead into a Beyonce song. Shame.
12. (Original post by PhysicsKid)
When x>2, all those expressions have a value greater than or equal to 0, so are not reflectes in the x-axis. Therefore the modulus graph of y=x-2 when x>2 has equation y=x-2 and not y=-(x-2) or*2-x. This is true for all the expressions, so once you've stated your range considerations, you can effectively ignore the mod signs and solve.
Ahh right i understand now, but how would i go about solving the equation?

Thanks!
13. (Original post by BabyMaths)
If x>2 then x+1>3 and so |x+1|=x+1.
Ahh ok, but i still dont know how i could re write the equation, could you show me please so i can figure out what has happened?

Would it be (x+1)-x+3(x-1)-2(x-2)-(x+2)=0 ?
14. (Original post by Zenarthra)
Ahh right i understand now, but how would i go about solving the equation?

Thanks!
The mod signs serve as brackets, so expand, collect like terms and it's then just a case of solving a linear equation
15. You have to consider the regions in which the different signs act on

When x > 2, no mod signs are in effect

What about between 2 and 1? 1 and 0?

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16. (Original post by PhysicsKid)
The mod signs serve as brackets, so expand, collect like terms and it's then just a case of solving a linear equation
(x+1)-x+3(x-1)-2(x-2)-(x+2)=0
If i expand out and simplify dont i get x-6 = x+2?
17. (Original post by Arithmeticae)
You have to consider the regions in which the different signs act on

When x > 2, no mod signs are in effect

What about between 2 and 1? 1 and 0?

Posted from TSR Mobile
In those further region there are mods signs that act, but how does this help me solve it?
And also what is the effect of adding mod equations, ofr instance what would |x+1|-|x| look like?
Just a straight line but reflected at the part where it hits the x axis?

Thanks!
18. Nvm i got it now, ty.
19. (Original post by Zenarthra)
In those further region there are mods signs that act, but how does this help me solve it?
And also what is the effect of adding mod equations, ofr instance what would |x+1|-|x| look like?
Just a straight line but reflected at the part where it hits the x axis?

Thanks!
I would suggest forgetting about graphs and just follow the algebra through!

When x > 0 then |x| = x and |x + 1| = x + 1 so |x + 1| - |x| = x + 1 - x = 1

Now deal with the other regions similarly on a case by case basis.
20. (Original post by davros)
I would suggest forgetting about graphs and just follow the algebra through!

When x > 0 then |x| = x and |x + 1| = x + 1 so |x + 1| - |x| = x + 1 - x = 1

Now deal with the other regions similarly on a case by case basis.
Yeah i understand now, just out of curiosity what would this graph look like?

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