"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
$$x+4&gt;\frac{2}{x+3}$$
As I said: this is easy. Multiply both sides by $(x+3)^2$, take everything over to one side and then factorise. This gives $(x+2)(x+3)(x+5)&gt;0$ and so $-5&lt;x&lt;-3$ or $x&gt;-2$.

Part (b) [1 mark]

This is where it gets a little odd. The question says: Deduce the values of $x$ for which
$$x+4&gt;\frac{2}{|x+3|}$$
According to the solutions, part (b) should be blindingly obvious. The answer is $x&gt;-2$. 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. $x-1&gt;2 \iff x&gt;3$, while $|x-1|&gt;2 \iff x&lt;-1 \ \ \text{or} \ \ x&gt;3$.
The right-hand-side of (b) is positive. This forces $x &gt; -4$. On the other hand, if $-4 &lt; x &lt; -3$, the left-hand side is less than $1$ but the right-hand-side is greater than $1$. Thus $x &gt; -3$.

This means $|x + 3| = x+3$ so the two inequalities (a) and (b) are the same. Since the solution to (a) is $-5 &lt; x &lt; -3$ or $x &gt; -2$, you must have $x &gt; -2$.

I don't know if it is blindingly obvious, but there you go.
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 $x+3 &lt; 0$ then you've got $(x+4)(x+3) &lt; 2$ (you have to change the sign, because you multipy by a negative figure)
this case return the solution $−5&lt;x&lt;−3$

And the second case $x+3 \geq 0$ returns $x&gt;−2.$

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.