• Support PF! Buy your school textbooks, materials and every day products Here!

Help with inecuation

  • Thread starter inverse
  • Start date
  • #1
26
0
Hello!

a)[itex]\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

b)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}[/itex]

c)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

Can you tell me how I can resolve it?

In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2,

but when I pass 2x-1 to multipliy with 1 "(3<1·2x-1)" Why I have not change "<" by ">" The denominator is negative, when is negative I thought I had change it.

The solution set is [itex](-\infty,\frac{1}{2})\cup(2,+\infty)[/itex]
for x> 2 not to have to change the sign to move the denominator to multiply by 1.

Thank you very much for your help :)
 
Last edited:

Answers and Replies

  • #2
77
0
If x<-4/3 then (3x+1)/(3x+4) is always greater than one. When x=0, (3x+1)/(3x+4)=1/4. When x=infinity, (3x+1)/(3x+4)=1. There's some region, while x is positive, that the expression is within your region of interest.
 
  • #3
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998


Hello!

a)[itex]\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

b)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}[/itex]

c)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

Can you tell me how I can resolve it?

In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2,

but when I pass 2x-1 to multiply with 1 "(3<1·2x-1)" Why I have not change "<" by ">" The denominator is negative, when is negative I thought I had change it.

The solution set is [itex](-\infty,\frac{1}{2})\cup(2,+\infty)[/itex]
for x> 2 not to have to change the sign to move the denominator to multiply by 1.

Thank you very much for your help :)
What method have you been taught to solve rational inequalities?

I like to solve for equality & then use the fact that the rational function is continuous, except where the denominator is zero. Solving for equality eliminates the need to separate into cases depending upon the sign of the denominator.

Look at (c): [itex]\displaystyle \frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

Solve: [itex]\displaystyle \frac{1}{2}=\frac{3x+1}{3x+4}[/itex]

[itex]3x+4=6x+2[/itex]

[itex]\displaystyle x=\frac{2}{3}[/itex]

Solve: [itex]\displaystyle \frac{3x+1}{3x+4}=\frac{2}{3}[/itex]

[itex]9x+3=6x+8[/itex]

[itex]\displaystyle x=\frac{5}{3}[/itex]​

Check easy to use numbers: one to the left of 2/3 one between 2/3 & 5/3 and one to the right of 5/3 . Which of these fulfill the inequality?
 
  • #4
26
0
Oh I see, [itex]\frac{2}{3}<x<\frac{5}{3}[/itex]

Can you help me with a and b?

Thank you
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998
Oh I see, [itex]\frac{2}{3}<x<\frac{5}{3}[/itex]

Can you help me with a and b?

Thank you
Can you follow the method I used for part (c) ?

Post your efforts for (a) & (b).

(There are other methods, of course!)

Is there a typo in your posted expression for (a)?

[itex]\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

or

[itex]\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}[/itex]
 
  • #6
26
0
Sorry, c) is [itex]\frac{3}{2x-1}<1[/itex]

So I know that x must be less than 1/2 for which the inequality. In the denominator of 3/2x-1 "I superimpose" if x takes values ​​(-infinity, 1/2) x is negative, thus passing the denominator to multiply with 1 if you change the "<" by "> ".
Then 3> 2x-1, 4> 2x and 2> x

Why is x>2? and not 2>x, What am I doing wrong?
 
  • #7
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998
Sorry, c) is [itex]\frac{3}{2x-1}<1[/itex]

So I know that x must be less than 1/2 for which the inequality. In the denominator of 3/2x-1 "I superimpose" if x takes values ​​(-infinity, 1/2) x is negative, thus passing the denominator to multiply with 1 if you change the "<" by "> ".
Then 3> 2x-1, 4> 2x and 2> x

Why is x>2? and not 2>x, What am I doing wrong?
Multiplying an inequality by a negative number reverses the direction of the inequality.


If x < 1/2, then 3/(2x-1) is negative, so it's definitely less than 1.

What if [itex]\displaystyle 2x-1>0\ \text{ i.e. }\ x>\frac{1}{2}\ ?[/itex]

Then multiplying [itex]\displaystyle \frac{3}{2x-1}<1[/itex] by (2x-1) gives:

[itex]\displaystyle 3< 2x-1[/itex]

Solving for x gives: [itex]\displaystyle x>\frac{3}{2}\ .[/itex]

Put the two answers together.
 
  • #8
26
0
I already understand c), but I still don't understand a) and b)
 
  • #9
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,302
998
Is there a typo in your posted expression for (a)?

[itex]\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

or

[itex]\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}[/itex]

or

something else?
 
  • #10
26
0
not only this, but I already know to solve this, thanks anyway
 
Top