4.6 Inequalities
An inequality is solved much like an equation, with one crucial difference.
Note 4.28. Multiplying or dividing both sides by a negative number reverses the inequality sign. From \(-2x<6\) we get \(x>-3\), not \(x<-3\).
That single rule is also why a rational inequality cannot be solved by cross-multiplying: the sign of the denominator is unknown, so we cannot tell whether to flip the sign. The safe method is to make one side zero and examine signs.
Solution. Factorise. \(x^2-x-6=(x-3)(x+2)\), so the critical values are \(x=-2\) and \(x=3\). These split the number line into three intervals.
Test the sign in each.
| Interval | \(x<-2\) | \(-2<x<3\) | \(x>3\) |
| \(x+2\) | \(-\) | \(+\) | \(+\) |
| \(x-3\) | \(-\) | \(-\) | \(+\) |
| product | \(+\) | \(-\) | \(+\) |
We want the product positive, so \[x<-2\quad \text {or}\quad x>3,\qquad \text {that is}\qquad (-\infty ,-2)\cup (3,\infty ).\] The inequality is strict, so both endpoints are excluded.
Solution. Do not cross-multiply, since \(x+5\) may be negative. Bring everything to one side instead: \[\frac {8}{x+5}-4<0\quad \implies \quad \frac {8-4(x+5)}{x+5}<0 \quad \implies \quad \frac {-4x-12}{x+5}<0.\] Factor the numerator: \[\frac {-4(x+3)}{x+5}<0.\] Dividing both sides by \(-4\) flips the sign: \[\frac {x+3}{x+5}>0.\]
The critical values are \(x=-5\), where the expression is undefined, and \(x=-3\), where it is zero.
| Interval | \(x<-5\) | \(-5<x<-3\) | \(x>-3\) |
| \(x+3\) | \(-\) | \(-\) | \(+\) |
| \(x+5\) | \(-\) | \(+\) | \(+\) |
| quotient | \(+\) | \(-\) | \(+\) |
\[\therefore \quad x<-5\quad \text {or}\quad x>-3,\qquad (-\infty ,-5)\cup (-3,\infty ).\]
Remark 4.31. \(x=-5\) is excluded because the expression is undefined there, and \(x=-3\) because the inequality is strict. A value that makes a denominator zero is never part of the solution, whatever the inequality sign — which is a different reason from the one that excludes \(-3\).
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