Solution set of $\frac x{x+2}>0\land\frac{x+1}{x+2}<1$

Question

Okay people, $\Im$ stupid. I have been working on this for literal 2 days but still I can't understand.

Here is the question: $$\frac x{x+2}>0\land\frac{x+1}{x+2}<1$$ What's the solution set? [Answer is $(0,\infty)$]

I found first inequality means $x > 0$ and second one says $x > -2$ but that doesn't make the solution set given by my lecturer which is $(0,\infty)$.

How did you get to those answers? Your lecturer is correct, so it would help if you could explain your reasoning. — Toby Mak, Nov 23 '20 at 11:38
First one is rather simple just multiple both sides with (x+2) and you will get x > 0. For second one substract 1 from both sides after simplifying u will get -1/(x+2)<0 multiply both sides with +1 and u will get 1/(x+2)>0 which basicly means x+2>0, x=/-2 and x is greater than -2 (x > -2). — UPiynar, Nov 23 '20 at 11:42
How can you do that? Like you are cancelling the (-2,0) area how? @player3236 — UPiynar, Nov 23 '20 at 11:44
If $x \in (-2,0)$, $x > -2$ is satisfied but not $x > 0$. If $x > 0$, $x > -2$ is automatically satisfied. — player3236, Nov 23 '20 at 11:47

score 1 · Accepted Answer · answered Nov 23 '20 at 11:53

1

The issue with your reasoning is when you multiply by a negative number, the inequality sign changes. Therefore, it is not true that $x > 0$ for all real $x$, but only when $x + 2 > 0$.

For the first part, I recommend that you split into cases. When $x + 2 > 0$, you do get $x > 0$. But when $x + 2 < 0$, then multiplying by $x+2$ on both sides gives:

$$x \color{red}{<} x+2 $$

which is true for all $x$ in the condition. Therefore, the possible values of $x$ are $x > 0, x < -2$.

For the second part, $-\frac{1}{x+2} < 0$ is correct so you can continue. From here, multiply by $-1$ to get:

$$\frac{1}{x+2} \color{red}{>} 0$$

and now use a similar method to find the possible values of $x$.

answered Nov 23 '20 at 11:53

Toby Mak

16,827

It says I shouldn't use comments for thanking but thank you so much. – UPiynar Nov 23 '20 at 12:16
No problem! I'm glad I could have helped. – Toby Mak Nov 23 '20 at 12:20
Regarding "thank you" in comments, most users will appreciate a thank you, as long as it's not just "thank you", but rather "thank you for helping" or "thank you, I can now solve the problem". For more information, see this Maths Meta SE thread about thanking in comments. – Toby Mak Nov 23 '20 at 12:22

score 0 · Answer 2 · answered Nov 23 '20 at 16:57

0

The best way to solve this kind of inequalities is not to split to different cases but to $\underline{\text{combine the fractions}}$.

For your first inequality: $$\frac{x}{x+2} >0 \iff x(x+2)>0 \iff x \in (-\infty, -2)\cup (0, \infty) \tag 1$$

For your second inequality: $$\frac{x+1}{x+2} < 1 \iff \frac{x+1}{x+2}-1 = - \frac{1}{x+2} < 0 \iff x+2 >0 \tag 2$$

Combine (1) and (2) you get $x>0$.

For another example, see Solving basic inequality

answered Nov 23 '20 at 16:57

Neat Math

4,790

Thats a different way to look at this, I will practice with this method too. Thanks for ur help. – UPiynar Nov 23 '20 at 20:15
Try $\frac {2}{x-2} - \frac{2}{x+3} > -1.$ – Neat Math Nov 23 '20 at 20:20
I found the solution set (x<-3 x>2)which is correct (I checked from Wolfram) but I would like to see your solution if you have time. – UPiynar Nov 23 '20 at 21:51
Just move $-1$ to the left and combine the fractions to get $\frac{x^2+x+4}{(x-2)(x+3)}>0$. – Neat Math Nov 23 '20 at 22:19

score 0 · Answer 3 · answered Nov 28 '20 at 17:30

0

ok so lets first consider the first inequality: $$\frac{x}{x+2}>0\tag{1}$$ for this to be true either $x>0$ so the top and bottom are both positive, or we can have $x<-2$ and so the solution for this inequality would be $x\in(-\infty,-2)\wedge(0,\infty)$.

Now for the second: $$\frac{x+1}{x+2}<1\tag{2}$$ $$1-\frac{1}{x+2}<1$$ $$-\frac 1{x+2}<0$$ $$\frac{1}{x+2}>0$$ and it is clear from this that the solution is $x>-2$ and so: $x\in(-2,\infty)$. Now for both to be simultaneously true we need to find where these domains overlap, which would be $x\in(0,\infty)$

answered Nov 28 '20 at 17:30

Henry Lee

12,215

Try plotting both functions graphically and shading the regions where the inequality is true, the overlap will represent your solution – Henry Lee Nov 28 '20 at 17:32
https://www.desmos.com/calculator/vcn4zegzdt – Henry Lee Nov 28 '20 at 17:33

Solution set of $\frac x{x+2}>0\land\frac{x+1}{x+2}<1$

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