need someone to check my work

[abel muroi]

I was given a problem with the non linear inequality x(2 -3x) <= 0 and was asked to put the solution in interval notation

I used the method where i had to get two values from the inequality and put them on the number line so i can check numbers from each interval

the two values i got is 0 and 2/3.

This is how i got my two values

i split the inequality into two "equations"

2 - 3x <= 0

3x <= -2

x >= 2/3 ((i dont think the greater than sign really matters here since i was already planning on testing the fraction on a number line))

and the second equation was just x <= 0


so my solution is

(negative infinity, 0 ) u ( 2/3, positive infinity)

of course, i used the symbol that represents infinity but i cant put it here

did i do the problem correctly?

 

[Ishuda]

I was given a problem with the non linear inequality x(2 -3x) <= 0 and was asked to put the solution in interval notation

I used the method where i had to get two values from the inequality and put them on the number line so i can check numbers from each interval

the two values i got is 0 and 2/3.

This is how i got my two values

i split the inequality into two "equations"

2 - 3x <= 0

3x <= -2

x >= 2/3 ((i dont think the greater than sign really matters here since i was already planning on testing the fraction on a number line))

and the second equation was just x <= 0


so my solution is

(negative infinity, 0 ) u ( 2/3, positive infinity)

of course, i used the symbol that represents infinity but i cant put it here

did i do the problem correctly?

Some of your actual explanations needed some improvement but you got the right answer for the somewhat right reasons I believe (assuming you did the number testing on the number line). Oh, and that should be semi-closed intervals, i.e.
(negative infinity, 0 ] u [ 2/3, positive infinity)
meaning 0 and 2.3 are included.

To go through what I would consider a (more) proper explanation for this type approach:
Find where
x(2 -3x) <= 0
First: Find where
x (2 -3x) = 0.
This gives two places, x = 0 and x=2/3. Mark them on the number line. That divides the number line into three parts we need to test:
1: x < 0. Choose x = -10.
(-10) * (2 - 3 * (-10)) = (-10) * (32) = -320 < 0
2: 0 < x < 2/3. Choose x = 1/3.
(2/3) * ( 2 - 3 * (1/3) ) = (1/3) * (1) = 1/3 > 0
3: x > 2/3. Choose x = 10.
10 * (2 - 3 * 10) = 10 * (-28) = -280 < 0.
Since it is less than or equal to zero we include 0 and 2/3, so we have
(infinity, 0] u [2/3, infinity)

 

[Steven G]

I was given a problem with the non linear inequality x(2 -3x) <= 0 and was asked to put the solution in interval notation

I used the method where i had to get two values from the inequality and put them on the number line so i can check numbers from each interval

the two values i got is 0 and 2/3.

This is how i got my two values

i split the inequality into two "equations"

2 - 3x <= 0

3x <= -2

x >= 2/3 ((i dont think the greater than sign really matters here since i was already planning on testing the fraction on a number line))

and the second equation was just x <= 0


so my solution is

(negative infinity, 0 ) u ( 2/3, positive infinity)

of course, i used the symbol that represents infinity but i cant put it here

did i do the problem correctly?

Abel, all looks fine. As Ishuda pointed out it is really difficult to really know if you got the correct answer for the right reason without seeing your work.
Please provide your work so we can be happy to see that you did everything correct or to further guide you if you made a mistake. Good luck!

 

[Ishuda]

...did i do the problem correctly?

Now, having said all that above, I do believe that doing it the way it has been presented here in a couple of threads is a better way to go because you have a better understanding of the problem (and, then again, may you personally don't, so take your choice). The important thing to notice in this situation is that if the product of two factors are negative then we have two cases:

1. The first factor is negative and the second factor is positive. For our case that means x<0 AND 2-3x>0 which translate to x<0 AND x<2/3. x < 0 covers both situations and since it is a less that or equal to condition, x = 0 is included.

2. The first factor is positive and the second factor is negative. For our case that means x>0 AND 2-3x<0 which translate to x>0 AND x>2/3. x > 2/3 covers both situations and since it is a less that or equal to condition, x = 2/3 is included.

Putting the two cases together we get the same answer (which is always very nice)
(infinity, 0] u [2/3, infinity)

 

[Steven G]

Abel,
Each and every one who offered you help with these problems have suggested that you do it the way you were advised to do it. There must be a reason why we all say this. Please reconsider what we are suggesting.

 

[abel muroi]

Some of your actual explanations needed some improvement but you got the right answer for the somewhat right reasons I believe (assuming you did the number testing on the number line). Oh, and that should be semi-closed intervals, i.e.
(negative infinity, 0 ] u [ 2/3, positive infinity)
meaning 0 and 2.3 are included.

To go through what I would consider a (more) proper explanation for this type approach:
Find where
x(2 -3x) <= 0
First: Find where
x (2 -3x) = 0.
This gives two places, x = 0 and x=2/3. Mark them on the number line. That divides the number line into three parts we need to test:
1: x < 0. Choose x = -10.
(-10) * (2 - 3 * (-10)) = (-10) * (32) = -320 < 0
2: 0 < x < 2/3. Choose x = 1/3.
(2/3) * ( 2 - 3 * (1/3) ) = (1/3) * (1) = 1/3 > 0
3: x > 2/3. Choose x = 10.
10 * (2 - 3 * 10) = 10 * (-28) = -280 < 0.
Since it is less than or equal to zero we include 0 and 2/3, so we have
(infinity, 0] u [2/3, infinity)

What you just wrote was exactly what I did to solve the problem. Like I said before, i got two places (values) from the inequality, which were 0 and 2/3. Then I put them on the number line which divided the number line into 3 parts. Then i choose a random number from each side and tested those numbers by substituting them for x in the inequality.

So i guess the only mistake that i did was forgetting to use the semi closed intervals

 

[abel muroi]

Abel,
Each and every one who offered you help with these problems have suggested that you do it the way you were advised to do it. There must be a reason why we all say this. Please reconsider what we are suggesting.

I thought you said that i should stick with methods that i will remember for the rest of my life, the method that ishuda just showed me is the method that i prefer. The other methods just seem to confuse me.

 

[Steven G]

I thought you said that i should stick with methods that i will remember for the rest of my life, the method that ishuda just showed me is the method that i prefer. The other methods just seem to confuse me.

Just because you remember it now does not mean you will remember it in 5 years from now.

If our method is confusing you then this is a problem. There are only a few reasons why this method confuses you. Maybe you do not know that for a product of two numbers to be positive the numbers must have the same sign (both positive or both negative). Possibly you do not know that if a product of two numbers is negative then one number is positive and the other number is negative. Since you are writing in english I will assume that you do your work in english. The last possible problem you can have is that you do not know the difference between AND and OR.
To succeed in algebra you MUST know the three areas mentioned in the above paragraph. This is not negotiable. Not knowing the above will severely affect your grade. Now once you know the above then you will like the method we use.

 

[ClaireWu]

I was given a problem with the non linear inequality x(2 -3x) <= 0 and was asked to put the solution in interval notation

I used the method where i had to get two values from the inequality and put them on the number line so i can check numbers from each interval

the two values i got is 0 and 2/3.

This is how i got my two values

i split the inequality into two "equations"

2 - 3x <= 0

3x <= -2

x >= 2/3 ((i dont think the greater than sign really matters here since i was already planning on testing the fraction on a number line))

and the second equation was just x <= 0


so my solution is

(negative infinity, 0 ) u ( 2/3, positive infinity)

of course, i used the symbol that represents infinity but i cant put it here

did i do the problem correctly?

My way of solving this problem :

x(2-3x) <= 0

-x(3x-2) <= 0

x(3x-2) >= 0

Using a number line and testing the 3 intervals marked by x=0 and x=2/3
the answer is {-infinity < x <= 0} U {2/3 <= x < infinity}