Solve 4x-5/x-3leq(1/2)

$$\frac { 4 x - 5 } { x - 3 } \leq ( \frac { 1 } { 2 } )$$

Solve for x

$$x\in [1,3)$$

Solution Steps

Denominator x-3 cannot be zero since division by zero is not defined. There are two cases.

$$x-3>0 x-3<0$$

Consider the case when x-3 is positive. Move -3 to the right hand side.

$$x>3$$

The initial inequality does not change the direction when multiplied by x-3 for x-3>0.

$$4x-5\leq \frac{1}{2}\left(x-3\right)$$

Multiply out the right hand side.

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

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

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

Combine like terms.

$$\frac{7}{2}x\leq \frac{7}{2}$$

Divide both sides by \frac{7}{2}. Since \frac{7}{2} is positive, the inequality direction remains the same.

$$x\leq 1$$

Consider condition x>3 specified above.

$$x\in \emptyset$$

Now consider the case when x-3 is negative. Move -3 to the right hand side.

$$x<3$$

The initial inequality changes the direction when multiplied by x-3 for x-3<0.

$$4x-5\geq \frac{1}{2}\left(x-3\right)$$

Multiply out the right hand side.

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

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

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

Combine like terms.

$$\frac{7}{2}x\geq \frac{7}{2}$$

Divide both sides by \frac{7}{2}. Since \frac{7}{2} is positive, the inequality direction remains the same.

$$x\geq 1$$

Consider condition x<3 specified above.

$$x\in [1,3)$$

The final solution is the union of the obtained solutions.

$$x\in [1,3)$$