Solve for x
x\in \left(-\infty,-1\right)\cup \left(2,\infty\right)
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x+1>0 x+1<0
Denominator x+1 cannot be zero since division by zero is not defined. There are two cases.
x>-1
Consider the case when x+1 is positive. Move 1 to the right hand side.
2x-1>x+1
The initial inequality does not change the direction when multiplied by x+1 for x+1>0.
2x-x>1+1
Move the terms containing x to the left hand side and all other terms to the right hand side.
x>2
Combine like terms.
x>2
Consider condition x>-1 specified above. The result remains the same.
x<-1
Now consider the case when x+1 is negative. Move 1 to the right hand side.
2x-1<x+1
The initial inequality changes the direction when multiplied by x+1 for x+1<0.
2x-x<1+1
Move the terms containing x to the left hand side and all other terms to the right hand side.
x<2
Combine like terms.
x<-1
Consider condition x<-1 specified above.
x\in \left(-\infty,-1\right)\cup \left(2,\infty\right)
The final solution is the union of the obtained solutions.
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