Solve the Inequality: ( (3x) / (4 - x) ) > 3x

J

Juxtaposition109

New member
Joined
Feb 15, 2016
Messages
8
It's supposed to be a greater than or equal to sign, but I'm not sure how to type that.

[IMG]http://www5b.wolframalpha.com/Calculate/MSP/MSP13711be3e7h22df74b26000036a6hc0ig5g5g9db?MSPStoreType=image/gif&s=28[/IMG]3 x">
 
Last edited by a moderator:
Juxtaposition109 said:
It's supposed to be a greater than or equal to sign, but I'm not sure how to type that.

[IMG]http://www5b.wolframalpha.com/Calculate/MSP/MSP13711be3e7h22df74b26000036a6hc0ig5g5g9db?MSPStoreType=image/gif&s=28[/IMG]3 x">
Click to expand...
I'm sorry, but Wolfram Alpha is saying that the image at the link does not exist. To learn how to post math as text, please review the illustrations here.

When you reply, please include a clear statement of your thoughts and efforts so far. Thank you! ;)
 
The problem is

((3x)/(4-x))>3x

I usually go ahead and multiply the entire problem by 4-x to knock out the fraction. After this I end up w/ 3x > 12x-3x^2.

I then bring subtract the 3x from both sides and end up w/ -3x^2-9x<0. I know I can factor out a -3x and the problem becomes -3x(x+9)<0. It's here that I'm stuck. I was also told that 4-x is a zero and I need to consider that as part of the solution.

Thx
 
Juxtaposition109 said:
The problem is

((3x)/(4-x))>3x
Click to expand...
I will guess that the instructions said something like "solve and check" or "find the solution interval and express (in some specified format)".

Juxtaposition109 said:
I usually go ahead and multiply the entire problem by 4-x to knock out the fraction.
Click to expand...
Wow. :shock:

A very important point (that your class apparently skipped over) is that one cannot multiply an inequality by a negative without also flipping the inequality sign. So, unless you broke this into two cases (one for x > 4 and another for x < 4), the results have to be wrong.

You can learn the correct method for this sort of exercise here. After you've read at least two lessons from the listing, please attempt the exercise, starting from:

. . . . .\(\displaystyle \dfrac{3x}{4\, -\, x}\, >\, 3x\)

. . .\(\displaystyle \mbox{Note: Not defined for }\, x\, =\, 4!\)

. . .\(\displaystyle 4\, -\, x\, >\, 0\, \mbox{ for }\, 4\, >\, x\, \mbox{ or }\, x\, <\, 4\)

. . .\(\displaystyle x\, \geq\, 4:\)

. . . . .\(\displaystyle 3x\, <\, 3x\, (4\, -\, x)\)

...and so forth. If you get stuck, please reply showing all of your work, starting from the above. Thank you! ;)
 
Juxtaposition109 said:
The problem is
((3x)/(4-x))>3x
I usually go ahead and multiply the entire problem by 4-x to knock out the fraction. PLEASE DO NOT!
Click to expand...
Your problem is equivalent to:
\(\displaystyle \dfrac{3x}{4-x}-3x\ge 0\) or simplified \(\displaystyle \dfrac{3x(x-3)}{x-4}\le 0\) You need to verify that.

The advantage there is that the are just three critical values: \(\displaystyle 0,~3,~4\).
 
Last edited:
pka said:
Your problem is equivalent to:
\(\displaystyle \dfrac{3x}{4-x}-3x\ge 0\) or simplified \(\displaystyle \dfrac{3x(x-3)}{x-4}\le 0\) You need to verify that.

The advantage there is that the are just three critical values: \(\displaystyle 0,~3,~4\).
Click to expand...




Thank you for the help :)
 
stapel said:
I will guess that the instructions said something like "solve and check" or "find the solution interval and express (in some specified format)".


Wow. :shock:

A very important point (that your class apparently skipped over) is that one cannot multiply an inequality by a negative without also flipping the inequality sign. So, unless you broke this into two cases (one for x > 4 and another for x < 4), the results have to be wrong.

You can learn the correct method for this sort of exercise here. After you've read at least two lessons from the listing, please attempt the exercise, starting from:

. . . . .\(\displaystyle \dfrac{3x}{4\, -\, x}\, >\, 3x\)

. . .\(\displaystyle \mbox{Note: Not defined for }\, x\, =\, 4!\)

. . .\(\displaystyle 4\, -\, x\, >\, 0\, \mbox{ for }\, 4\, >\, x\, \mbox{ or }\, x\, <\, 4\)

. . .\(\displaystyle x\, \geq\, 4:\)

. . . . .\(\displaystyle 3x\, <\, 3x\, (4\, -\, x)\)

...and so forth. If you get stuck, please reply showing all of your work, starting from the above. Thank you! ;)
Click to expand...

Is it really not defined for 4! = 24? :p
 
Juxtaposition109 said:
It's supposed to be a greater than or equal to sign, but I'm not sure how to type that.

[IMG]http://www5b.wolframalpha.com/Calculate/MSP/MSP13711be3e7h22df74b26000036a6hc0ig5g5g9db?MSPStoreType=image/gif&s=28[/IMG]3 x">
Click to expand...
If x>0 then the problem reduces to 1/(4-x)>1 which means 4-x< 1 which gives x>3. So what is the answer for this part??

If x<0 then the problem reduces to 1/(4-x)<1 which means 4-x>1 which gives x<3. So what is the answer for this part?

Can x=4??
 
You must log in or register to reply here.
Top