Solve 2x-3/x+5geq1

$$\frac{ 2x-3 }{ x+5 } \geq 1$$

Solve for x

$$x\in (-\infty,-5)\cup [8,\infty)$$

Solution Steps

Denominator x+5 cannot be zero since division by zero is not defined. There are two cases.

$$x+5>0 x+5<0$$

Consider the case when x+5 is positive. Move 5 to the right hand side.

$$x>-5$$

The initial inequality does not change the direction when multiplied by x+5 for x+5>0.

$$2x-3\geq x+5$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-x\geq 3+5$$

Combine like terms.

$$x\geq 8$$

Now consider the case when x+5 is negative. Move 5 to the right hand side.

$$x<-5$$

The initial inequality changes the direction when multiplied by x+5 for x+5<0.

$$2x-3\leq x+5$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$2x-x\leq 3+5$$

Combine like terms.

$$x\leq 8$$

Consider condition x<-5 specified above.

$$x<-5$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-5)\cup [8,\infty)$$