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education/math/Rational Inequalities.md
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@ -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
$$ \frac{2x-17}{x-5}
$$ \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.