nidantasnu

2022-07-10

For what x is $\frac{1}{x+3}>\frac{2}{x-3}$?

Jayvion Tyler

Expert

Step 1
Note that $x=-3$ and $x=3$ are not solutions since one or the other side of the inequality is undefined for these values of x.
Split the remaining possibilities into cases:
Case $x\in \left(-\infty ,-3\right)\cup \left(3,\infty \right)$
$x+3$ and $x-3$ are both negative or both positive.
In either case $\left(x+3\right)\left(x-3\right)>0$ so multiply both sides of the inequality by that to get:
$x-3>2x+6$
Subtract $x+6$ from both sides to get:
$-9>x$
Combining with the conditions of the case, that gives us the solution:
$x\in \left(-\infty ,-9\right)$
Case $x\in \left(-3,3\right)$.
$x+3>0$ and $x-3<0$, so $\left(x+3\right)\left(x-3\right)<0$, so multiply both sides of the inequality by this and reverse the inequality to get:
$x-3<2x+6$
Step 2
Subtract $x+6$ from both sides to get:
$-9
Combining with the conditions of the case, that gives us the solution:
$x\in \left(-3,3\right)$
Conclusion
Putting the two cases together, the original inequality holds for all
$x\in \left(-\infty ,-9\right)\cup \left(-3,3\right)$

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