Solve for x
x\in (-\infty,-3)\cup [-\frac{11}{5},\infty)
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x+3>0 x+3<0
Denominator x+3 cannot be zero since division by zero is not defined. There are two cases.
x>-3
Consider the case when x+3 is positive. Move 3 to the right hand side.
1-x\leq 4\left(x+3\right)
The initial inequality does not change the direction when multiplied by x+3 for x+3>0.
1-x\leq 4x+12
Multiply out the right hand side.
-x-4x\leq -1+12
Move the terms containing x to the left hand side and all other terms to the right hand side.
-5x\leq 11
Combine like terms.
x\geq -\frac{11}{5}
Divide both sides by -5. Since -5 is negative, the inequality direction is changed.
x<-3
Now consider the case when x+3 is negative. Move 3 to the right hand side.
1-x\geq 4\left(x+3\right)
The initial inequality changes the direction when multiplied by x+3 for x+3<0.
1-x\geq 4x+12
Multiply out the right hand side.
-x-4x\geq -1+12
Move the terms containing x to the left hand side and all other terms to the right hand side.
-5x\geq 11
Combine like terms.
x\leq -\frac{11}{5}
Divide both sides by -5. Since -5 is negative, the inequality direction is changed.
x<-3
Consider condition x<-3 specified above.
x\in (-\infty,-3)\cup [-\frac{11}{5},\infty)
The final solution is the union of the obtained solutions.
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