Math

[gillgill]

1) │3x-2│<= x+1 ; x>=-1
Case1: 3x-2>=0
x>= 2/3
3x-2<=x+1
x<=3/2

what is case 2???

2) │2-3x│ < 3x-4

3) │x-3│=x-2
How do u solve these?

[dextercioby]

Apply the definition of the absolute value correctly.
Case 1.3x-2>=0 (therefore x>=2/3) -------->3x-2<=x+1------------>x<=3/2.The solution: x\in [\frac{2}{3},\frac{3}{2}]

Case 2.3x-2<0 (therefore x<2/3)---------->-(3x-2)<=x+1------------>x>=1/4.The solution is: x\in [\frac{1}{4},\frac{2}{3})

The solution of the problem is found by reuniting the partial solutions
x\in [\frac{1}{4},\frac{2}{3}]

Do the same for the other 2...

Daniel.

[gillgill]

how do u know case 2 would be -(3x-2)<=x+1 with a negative sign?

[dextercioby]

Because that's the definition of the absolute value
|x|=x,for x>=0 and -x for x<0...

Daniel.

[gillgill]

hm.....can u do one more for me?

[dextercioby]

Nope.What is the result of applying the definition of an absolute value to point b)...?

Daniel.

[gillgill]

What point b?

[dextercioby]

Exercise 2),sorry...

Daniel.

[gillgill]

hm...ok...let me try
case 1: 2-3x>=0
x<=2/3

[learningphysics]

Apply the definition of the absolute value correctly.
Case 1.3x-2>=0 (therefore x>=2/3) -------->3x-2<=x+1------------>x<=3/2.The solution: x\in [\frac{2}{3},\frac{3}{2}]

Case 2.3x-2<0 (therefore x<2/3)---------->-(3x-2)<=x+1------------>x>=1/4.The solution is: x\in [\frac{1}{4},\frac{3}{2})


I believe the above should be:
x\in [\frac{1}{4},\frac{2}{3})

And the final solution, union:
x\in [\frac{1}{4},\frac{3}{2}]

[learningphysics]

I don't think you need to consider the value inside the || separately for positive and negative. (Unless your teacher wants you to do it that way to fully understand the steps.)

For the first question, I'd just say:
│3x-2│<= x+1 so
-(x+1)<=3x-2<=x+1
or in other words
-(x+1)<=3x-2 AND 3x-2<=x+1
The first inequality gives x>=1/4, the second gives x<=3/2

The solution is the intersection of x>=1/4 and x<=3/2, so the solution is:
1/4<=x<=3/2

Although it is instructive to consider separately the positive and negative values inside ||, it isn't necessary to solve the inequality.

[primarygun]

Do you mean consider the value of x is bigger or equal to 2/3 first? or Less than or equal ?
In the past, I have tried inequalities that with several absolute sign inside. It is extremely important to define the value first.
However, I haven't learnt this in my lessons. Maybe later. Therefore, I don't know whether in this question this distinction is needed.

[learningphysics]

I don't see why it is necessary...

if |a|<b, then

we know that a<b and a>-b

This is true whether or not a is positive or negative or 0.

If |a|>b, then

we know that a>b or a<-b. This statement is also true whether or not a is positive or negative or zero.

[dextercioby]

I believe the above should be:
x\in [\frac{1}{4},\frac{2}{3})

And the final solution, union:
x\in [\frac{1}{4},\frac{3}{2}]

Yes,thank you for noticing.I edited my post and now it's "dandy"... :wink:

Daniel.

[primarygun]

Yes. It is not necessary for this case.