"Simple" rational inequality - How does one deduce this? I am a little puzzled by a past exam question. Part (a) is very easy and part (b) is easy too. However, according to the question, and the mark scheme, I should be able to deduce the answer to part (b) from my answer to part (a) without any further calculation. I am able to find the answer to part (b), but I have to draw some graphs and it is far from straight forward.
Part (a) [6 marks]
Find the set of values of $x$ for which
$$x+4>\frac{2}{x+3}$$
As I said: this is easy. Multiply both sides by $(x+3)^2$, take everything over to one side and then factorise. This gives $(x+2)(x+3)(x+5)>0$ and so $-5<x<-3$ or $x>-2$.
Part (b) [1 mark]
This is where it gets a little odd. The question says: Deduce the values of $x$ for which
$$x+4>\frac{2}{|x+3|}$$
According to the solutions, part (b) should be blindingly obvious. The answer is $x>-2$. However, I don't see how one can deduce this trivially. I drew the graph and got the right answer, but a 1 mark question means it is obvious and there is no need for calculation.
Adding a modulus usually makes a problem more difficult and adds solutions, e.g. $x-1>2 \iff x>3$, while $|x-1|>2 \iff x<-1 \ \ \text{or} \ \ x>3$.
 A: The right-hand-side of (b) is positive. This forces $x > -4$. On the other hand, if $-4 < x < -3$, the left-hand side is less than $1$ but the right-hand-side is greater than $1$. Thus $x > -3$.
This means $|x + 3| = x+3$ so the two inequalities (a) and (b) are the same. Since the solution to (a) is $-5 < x < -3$ or $x > -2$, you must have $x > -2$.
I don't know if it is blindingly obvious, but there you go.
A: I don't know what they expected you to do. But notice that 
$$x+4>\frac{2}{|x+3|}\geq\frac{2}{x+3}$$
This means that any solution of part (b) is a solution of part (a) and that discarding solutions is enough. It is not difficult to get rid of the interval $(-5,-3)$, but I think some argument has to be written in the exam.
A: You can see the result from the calc out of a). 
but not out of yours. 
I multiply first $(x+3)$. Then you have 2 cases. 
First case $x+3 < 0$ then you've got $(x+4)(x+3) < 2$ (you have to change the sign, because you multipy by a negative figure)
this case return the solution $−5<x<−3$
And the second case $x+3 \geq 0$ returns $x>−2.$
It IS quite obvious, that if you multiply by $|x+3|$ there is only the second case. So only the solution of the second case is correct.
