From 166e316cd1d8635b1d778e2ac261291565d5d1b6 Mon Sep 17 00:00:00 2001 From: zleyyij Date: Wed, 17 Jan 2024 10:25:34 -0700 Subject: [PATCH] vault backup: 2024-01-17 10:25:34 --- education/math/Rational Inequalities.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/education/math/Rational Inequalities.md b/education/math/Rational Inequalities.md index 235d24d..985835b 100644 --- a/education/math/Rational Inequalities.md +++ b/education/math/Rational Inequalities.md @@ -9,4 +9,5 @@ $$ \frac{(x+1)^2(x-1)}{(x+4)(x-3)} \le 0 $$ In this case, solving the bottom means that you'll have empty points on $x - 4$ and $x - 3$. Solving the top means that you'll have filled in points at $x = -1$ and $x = 1$. At this point the number line will be divided into chunks. You can pick an arbitrary number in each chunk and plug it in for $x$. This will let you figure out which parts of the range are valid. The result is written in the form of $(m, M) \cup (m, M)$. If the other side isn't zero (EG, $< 3$), you'll move everything to one side. You can do this by multiplying the right side by the denominator on the bottom of the fraction -$$ \frac{2x-17}{x-5} \ No newline at end of file +$$ \frac{2x-17}{x-5} \lt 3 $$ +In this example, you'd multiply the right side by $\frac{x-5}{x-5}$. Once that's done, you can subtract it over and combine. \ No newline at end of file