# Help with inecuation

Hello!

a)$\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}$

b)$\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}$

c)$\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}$

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 $(-\infty,\frac{1}{2})\cup(2,+\infty)$
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 :)

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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.

SammyS
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Hello!

a)$\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}$

b)$\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}$

c)$\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}$

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 $(-\infty,\frac{1}{2})\cup(2,+\infty)$
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): $\displaystyle \frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}$

Solve: $\displaystyle \frac{1}{2}=\frac{3x+1}{3x+4}$

$3x+4=6x+2$

$\displaystyle x=\frac{2}{3}$

Solve: $\displaystyle \frac{3x+1}{3x+4}=\frac{2}{3}$

$9x+3=6x+8$

$\displaystyle x=\frac{5}{3}$​

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?

Oh I see, $\frac{2}{3}<x<\frac{5}{3}$

Can you help me with a and b?

Thank you

SammyS
Staff Emeritus
Homework Helper
Gold Member
Oh I see, $\frac{2}{3}<x<\frac{5}{3}$

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)?

$\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}$

or

$\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}$

Sorry, c) is $\frac{3}{2x-1}<1$

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?

SammyS
Staff Emeritus
Homework Helper
Gold Member
Sorry, c) is $\frac{3}{2x-1}<1$

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 $\displaystyle 2x-1>0\ \text{ i.e. }\ x>\frac{1}{2}\ ?$

Then multiplying $\displaystyle \frac{3}{2x-1}<1$ by (2x-1) gives:

$\displaystyle 3< 2x-1$

Solving for x gives: $\displaystyle x>\frac{3}{2}\ .$

I already understand c), but I still don't understand a) and b)

SammyS
Staff Emeritus
Homework Helper
Gold Member
Is there a typo in your posted expression for (a)?

$\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}$

or

$\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}$

or

something else?

not only this, but I already know to solve this, thanks anyway