doubt inequations

[inverse]

Hello!


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


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

c)\(\displaystyle \frac{3}{2x-1}<1\)



Can you tell me how I can resolve it?

pd: further clarification, do not understand why in c) passing to multiply the denominator does not change the sign of the inequality.


Thank you very much for your help

 

[JeffM]

Hello!


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


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

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



Can you tell me how I can resolve it?

pd: further clarification, do not understand why in c) passing to multiply the denominator does not change the sign of the inequality.


Thank you very much for your help

Are these supposed to be steps in solving one problem? Or are these three different problems. What EXACTLY does your book say is the problem?

 

[inverse]

three are the three inequalities that I'm not agree with the right result.

 

[JeffM]

three are the three inequalities that I'm not agree with the right result.

It is impossible to tell you where you are going wrong if you do not show your work.

 

[inverse]

It is impossible to tell you where you are going wrong if you do not show your work.

c) 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

What am I doing wrong?

 

[JeffM]

c) 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

What am I doing wrong?

You do realize that what you wrote is virtually incomprehensible.

You only know initially that \(\displaystyle x \ne \dfrac{1}{2}\) because otherwise the denominator would be zero.

But you must consider what happens on both sides of 1/2. In other words, you have two potential cases, one where the denominator is negative and another where the denominator is positive. I'll do the negative case. Then you try the positive case.

\(\displaystyle x < \dfrac{1}{2} \implies 2x < 1 \implies 2x - 1 < 1 - 1 \implies 2x - 1 < 0 \implies \dfrac{1}{2x - 1} < 0 \implies \dfrac{3}{2x - 1} < 0 < 3.\)

\(\displaystyle In\ short,\ x < \dfrac{1}{2} \implies \dfrac{3}{2x - 1} < 3.\)

But there remains the other case. Let's see your result for that case.

 

[stapel]

c)\(\displaystyle \frac{3}{2x-1}<1\)

[I] do not understand why in c) passing to multiply the denominator does not change the sign of the inequality.

Unless you know that "2x - 1" is negative, how can you know that you need to flip the sign?

This, by the way, is why you need to solve this in two separate cases, one where you must flip the sign and one where you must not. :wink: