Solve for x
x\in (-\infty,-5)\cup [3,\infty)
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\frac{3-x}{x+5}\leq 0
Anything times zero gives zero.
3-x\geq 0 x+5<0
For the quotient to be ≤0, one of the values 3-x and x+5 has to be ≥0, the other has to be ≤0, and x+5 cannot be zero. Consider the case when 3-x\geq 0 and x+5 is negative.
x<-5
The solution satisfying both inequalities is x<-5.
3-x\leq 0 x+5>0
Consider the case when 3-x\leq 0 and x+5 is positive.
x\geq 3
The solution satisfying both inequalities is x\geq 3.
x<-5\text{; }x\geq 3
The final solution is the union of the obtained solutions.
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