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I would like to know if somebody knows how to properly divide one inequality by another, as a resolution method similar to when we divide one equality by another.

Take this as an example: $x^2 - y^2 \lt 8$ and $x + y \gt 3$.

I want to know if it is possible to factor the left hand side of the first equation, to multiply by $-1$ the second equation and divide both equation as to obtain: $y - x \gt -8/3$. I don't really know if the sign must be < or >.

Thanks!

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  • $\begingroup$ You seem to be aware that multiplying both sides of an inequality by a negative number reverses the direction of the inequality. A similar effect occurs if one divides both sides by a negative number. It's better to break down the argument into several small steps, rather than the "giant leap forward" you propose. $\endgroup$ – hardmath Aug 28 '14 at 3:22
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    $\begingroup$ It isn't that simple. For e.g try "dividing" $3<4$ with $1<2$. Even if everything is positive, it doesn't make sense. Though it will be true to say $0<a<b, 0<c<d \implies a/d < b/c$, which can be obtained by multiplying the inequalities and then dividing by $cd$, the reverse implication doesn't hold, so you will miss solutions. $\endgroup$ – Macavity Aug 28 '14 at 3:46
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Following the comment above, here are some small steps:

Let us graph the two inequalities (using a domain where the corresponding functions exist).

The first one holds when $y^2 > x^2 - 8$.

So we can graph $y = \sqrt{x^2 - 8}$ and color the points above it blue; but, since there are two square roots, we also consider $y = -\sqrt{x^2 - 8}$ and color the points below it yellow.

Next, graph the second inequality, $y > -x + 3$, by drawing the line $y = -x + 3$, and color the points above it red.

Now, anything that is colored purple (i.e., red and blue) or orange (i.e., red and yellow) satisfies the inequalities.

Does this picture correspond to your conjectured $y > x - 8/3$?

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Suppose $x^2-y^2 < 8$ and $x+y > 3$. If $y - x \le -\dfrac{8}{3}$, then $x-y \ge \dfrac{8}{3}$, and hence $x^2-y^2 = (x-y)(x+y) > \dfrac{8}{3} \cdot 3 = 8$, a contradiction since $x^2-y^2 < 8$.

Thus, any $(x,y)$ which satisfy both $x^2-y^2 < 8$ and $x+y > 3$ will also satisfy $y - x < -\dfrac{8}{3}$.

However, the converse is not true, that is if $(x,y)$ satisfy $y-x < -\dfrac{8}{3}$, then it is not necessarily true that $x^2-y^2 < 8$ and $x+y > 3$. See the other comments for counterexamples.

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