Solve 3-x/5x+2<1

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

Solve for x

$$x\in \left(-\infty,-\frac{2}{5}\right)\cup \left(\frac{1}{6},\infty\right)$$

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.

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

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

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

Combine like terms.

$$-6x<-1$$

Divide both sides by -6. Since -6 is negative, the inequality direction is changed.

$$x>\frac{1}{6}$$

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

$$x>\frac{1}{6}$$

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.

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

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

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

Combine like terms.

$$-6x>-1$$

Divide both sides by -6. Since -6 is negative, the inequality direction is changed.

$$x<\frac{1}{6}$$

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

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

The final solution is the union of the obtained solutions.

$$x\in \left(-\infty,-\frac{2}{5}\right)\cup \left(\frac{1}{6},\infty\right)$$