Solve 4x-3/5-2x>-1

$$\frac { 4 x - 3 } { 5 - 2 x } > - 1$$

Solve for x

$$x\in \left(-1,\frac{5}{2}\right)$$

Solution Steps

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

$$5-2x>0 5-2x<0$$

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

$$-2x>-5$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

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

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

$$4x-3>-\left(5-2x\right)$$

Multiply out the right hand side.

$$4x-3>-5+2x$$

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

$$4x-2x>3-5$$

Combine like terms.

$$2x>-2$$

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

$$x>-1$$

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

$$x\in \left(-1,\frac{5}{2}\right)$$

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

$$-2x<-5$$

Divide both sides by -2. Since -2 is negative, the inequality direction is changed.

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

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

$$4x-3<-\left(5-2x\right)$$

Multiply out the right hand side.

$$4x-3<-5+2x$$

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

$$4x-2x<3-5$$

Combine like terms.

$$2x<-2$$

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

$$x<-1$$

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

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in \left(-1,\frac{5}{2}\right)$$