When I do it, it never goes anywhere the first time because there are so many conditions and cases when x or y is positive or negative. The worked solutions suggested a "quadrant" based method but it is still confusing. Is there an even easier way or will a question like this naturally be difficult to solve?
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Can you post the worked solutions in your question? You will have to split into cases, so you will need to work methodically but it can be done. – Toby Mak Dec 06 '20 at 08:16
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1Can you please add complete hypothesis and what you want exactly? It is unclear to me what you are asking. – Marktmeister Dec 06 '20 at 08:16
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1Yes I have posted the worked solutions @TobyMak – user71207 Dec 06 '20 at 08:18
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@Marktmeister hypothesis? I just want an easy way to solve this inequality with as little cases as possible – user71207 Dec 06 '20 at 08:18
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If $xy>0$ (the first and third quadrant excluding the vertical and horizontal axis), multiply $xy$ to both sides of the inequality, we have $y > x$ in the first and the third quadrant.
If $xy < 0$ (the second and fourth quadrant excluding the vertical and horizontal axis), multiply $xy$ to both sides of the inequality, we have $y < x$. This is just the entire fourth quadrant.
Siong Thye Goh
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Thank you I was looking for something like this. However, how does $y<x$ imply the entire 4th quadrant? – user71207 Dec 06 '20 at 08:23
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$y<x$ and $xy<0$ would imply the fourth quadrant. We were considering the second and the fourth quadrant and we can draw the region $y<x$ easily and notice that it doesn't intersect with the second quadrant but it interesct with the entire $4$-th quadrant. – Siong Thye Goh Dec 06 '20 at 08:26
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What do you mean by intersecting? $y<x$ does satisfy the 2nd quadrant – user71207 Dec 06 '20 at 08:30
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in the second quadrant, $y >0$ and $x < 0$, we have $x < 0< y$, that is $x<y$. We can't have $x < y$ and $y<x$ simultaneously. You can view intersect as when you sketch the region, the region that overlaps each other. – Siong Thye Goh Dec 06 '20 at 08:32
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Multiply both sides of the inequality by $xy$.
If $xy > 0$, this means $x > 0$ and $y > 0$, or $x < 0$ and $y < 0$. We have $y > x$, so sketch the upper half of the 1st quadrant and the upper half of the 3rd quadrant.
If $xy < 0$, this means $x < 0$ and $y > 0$, or $x > 0$ and $y < 0$. This gives $y < x$, but this is violated when $x < 0$ and $y > 0$. Therefore sketch the entire 4th quadrant as both conditions are satisfied.
Toby Mak
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here are the worked solutions (as requested). I guess its relatively concise - is there an even better way?
user71207
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