Solve (2x-3)/(5-x)leq1

$$\frac { ( 2 x - 3 ) } { ( 5 - x ) } \leq 1$$

Solve for x

$$x\in (-\infty,\frac{8}{3}]\cup (5,\infty)$$

Solution Steps

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

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

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

$$-x>-5$$

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

$$x<5$$

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

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

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.

$$3x\leq 8$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x\leq \frac{8}{3}$$

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

$$-x<-5$$

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

$$x>5$$

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

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

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.

$$3x\geq 8$$

Divide both sides by 3. Since 3 is positive, the inequality direction remains the same.

$$x\geq \frac{8}{3}$$

Consider condition x>5 specified above.

$$x>5$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,\frac{8}{3}]\cup (5,\infty)$$