Solve 2x-3/5x+2geq2

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

Solve for x

$$x\in [-\frac{7}{8},-\frac{2}{5})$$

Solution Steps

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

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

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

$$5x>-2$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

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

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

$$2x-3\geq 2\left(5x+2\right)$$

Multiply out the right hand side.

$$2x-3\geq 10x+4$$

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

$$2x-10x\geq 3+4$$

Combine like terms.

$$-8x\geq 7$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x\leq -\frac{7}{8}$$

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

$$x\in \emptyset$$

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

$$5x<-2$$

Divide both sides by 5. Since 5 is positive, the inequality direction remains the same.

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

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

$$2x-3\leq 2\left(5x+2\right)$$

Multiply out the right hand side.

$$2x-3\leq 10x+4$$

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

$$2x-10x\leq 3+4$$

Combine like terms.

$$-8x\leq 7$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x\geq -\frac{7}{8}$$

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

$$x\in [-\frac{7}{8},-\frac{2}{5})$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{7}{8},-\frac{2}{5})$$