Solve 5x/4+xleq1

$$\frac { 5 x } { 4 + x } \leq 1$$

Solve for x

$$x\in (-4,1]$$

Solution Steps

Denominator 4+x cannot be zero since division by zero is not defined. There are two cases.

$$4+x>0 4+x<0$$

Consider the case when 4+x is positive. Move 4 to the right hand side.

$$x>-4$$

The initial inequality does not change the direction when multiplied by 4+x for 4+x>0.

$$5x\leq 4+x$$

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

$$5x-x\leq 4$$

Combine like terms.

$$4x\leq 4$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x\leq 1$$

Consider condition x>-4 specified above.

$$x\in (-4,1]$$

Now consider the case when 4+x is negative. Move 4 to the right hand side.

$$x<-4$$

The initial inequality changes the direction when multiplied by 4+x for 4+x<0.

$$5x\geq 4+x$$

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

$$5x-x\geq 4$$

Combine like terms.

$$4x\geq 4$$

Divide both sides by 4. Since 4 is positive, the inequality direction remains the same.

$$x\geq 1$$

Consider condition x<-4 specified above.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in (-4,1]$$