Solving inequalities: 4/(x + 1) <= 3/(x + 2)

[Bladesofhalo]

We skipped over this section a bit, but come finals we need to know. How would u proceed with solving this inequality?
4/(x + 1) is less than or equal to 3/( x + 2)

Am I proceeding in the right direction? Heres my work.
Multiplying both sides by (x + 1)(x + 2) yields:
4x + 8 is less than or equal to 3x + 1

But the answer is (negative infinity, -5) u (-2, -1)

Something I did wrong?

 

[stapel]

Multiplying both sides by (x + 1)(x + 2) yields...

When you multiply or divide through an inequality by a negative, you have to flip the inequality sign. On what basis did you not flip the sign?

Instead, try moving everything over to one side, converting to the common denominator, and finding the zeroes and vertical asymptotes of the resulting rational expression. These points will split the number line into intervals. Check the sign of the rational expression on each interval, and choose the appropriate intervals for your solution. :idea:

If you get stuck, please reply showing your work and reasoning. Thank you!

Eliz.

 

[Deleted member 4993]

Re: Solving inequalities: 4/(x + 1) <= 3/(x + 3)

We skipped over this section a bit, but come finals we need to know. How would u proceed with solving this inequality?
4/(x + 1) is less than or equal to 3/( x + 2)

Am I proceeding in the right direction? Heres my work.
Multiplying both sides by (x + 1)(x + 2) yields:
4x + 8 is less than or equal to 3x + 1

But the answer is (negative infinity, -5) u (-2, -1)

Something I did wrong?

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

This is done easiest if you invert (flip) both-sides changing the inequality sign

\(\displaystyle \frac{x+1}{4}\, >=\, \frac{x+2}{3}\)

Now multiply both sides by 12 and continue....

 

[Bladesofhalo]

Re: Solving inequalities: 4/(x + 1) <= 3/(x + 3)

So after doing this:
(12x + 12)/4 >= (12X + 24) / 3
Further simplified:
(3x + 4) >= (4x + 8)

Now shall I do what eliz stated and move everything to one side and calculate zeroes, asymptotes, and such?

 

[stapel]

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

This is done easiest if you invert (flip) both-sides changing the inequality sign

\(\displaystyle \frac{x+1}{4}\, >=\, \frac{x+2}{3}\)

Now multiply both sides by 12 and continue....

Unfortunately, the above yields exactly the same wrong answer as the poster had originally obtained.

By flipping the fractions, the vertical asymptotes were lost, and thus the other solution interval. Following the previously-listed steps will provide the correct answer, which matches the book's answer. :wink:

Eliz.

 

[Deleted member 4993]

Re:

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

This is done easiest if you invert (flip) both-sides changing the inequality sign

\(\displaystyle \frac{x+1}{4}\, >=\, \frac{x+2}{3}\)

Now multiply both sides by 12 and continue....

Unfortunately, the above yields exactly the same wrong answer as the poster had originally obtained.

By flipping the fractions, the vertical asymptotes were lost, and thus the other solution interval. Following the previously-listed steps will provide the correct answer, which matches the book's answer. :wink:

Eliz.

I stand corrected.

I violated my own cardinal rule - thou shalt always sketch function/s before concluding.