Solve x-3/2x+5<1

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

Solve for x

$$x\in \left(-\infty,-8\right)\cup \left(-\frac{5}{2},\infty\right)$$

Solution Steps

Denominator 2x+5 cannot be zero since division by zero is not defined. There are two cases.

$$2x+5>0 2x+5<0$$

Consider the case when 2x+5 is positive. Move 5 to the right hand side.

$$2x>-5$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x>-\frac{5}{2}$$

The initial inequality does not change the direction when multiplied by 2x+5 for 2x+5>0.

$$x-3<2x+5$$

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

$$x-2x<3+5$$

Combine like terms.

$$-x<8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x>-8$$

Consider condition x>-\frac{5}{2} specified above.

$$x>-\frac{5}{2}$$

Now consider the case when 2x+5 is negative. Move 5 to the right hand side.

$$2x<-5$$

Divide both sides by 2. Since 2 is positive, the inequality direction remains the same.

$$x<-\frac{5}{2}$$

The initial inequality changes the direction when multiplied by 2x+5 for 2x+5<0.

$$x-3>2x+5$$

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

$$x-2x>3+5$$

Combine like terms.

$$-x>8$$

Divide both sides by -1. Since -1 is negative, the inequality direction is changed.

$$x<-8$$

Consider condition x<-\frac{5}{2} specified above. The result remains the same.

$$x<-8$$

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-8\right)\cup \left(-\frac{5}{2},\infty\right)$$