# Am I solving this inequality correctly?

Say I have an inequality as such ($a,b$ and $c$ are all positive numbers)

$a > b \times c$

Now I need to move $c$ to the L.H.S and separate it from others so this inequality would become

$c < \frac{a}{b}$

Am I doing this right? Since I changed sides of $c$ I also changed the equality sign.

• It depends on signs of $a,b$ and $c$. – Zia Sep 12 '13 at 6:38

Yes, that is correct. You can see that by writing this: $$a > bc\\ \frac{a}{b} > c$$

### Edit

Important assumption was that $b > 0$. If $b < 0$ that itself switches the inequality and then switching sides is a second flip (so it stays the same).

For example. When $a = 6, b = 2, c = 1$ we have: $$6 > 2\cdot 1\\ 3 > 1\\ 1 < 3$$

But when $a = 6, b = -2, c = 1$ we have: $$6 > -2\cdot 1\\ -3 \mathbf{<} 1\\ 1 \mathbf{>} 3$$

$a > (b)(c) \to \frac{a}{b} > c$.. If this is just a regular inequality, I'm not sure what the question is... There is no need to flip the sign of equality.
If it was the case of $a > (-b)(c)$, we would have $\frac{a}{b} < c$
Plug in arbitrary values to see this is the case.