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