Solve 5+x/x+13+22geq4/9

$$\frac { 5 + x } { x + 13 + 22 } \geq \frac { 4 } { 9 }$$

Solve for x

$$x\in (-\infty,-35)\cup [19,\infty)$$

Solution Steps

Add 13 and 22 to get 35.

$$\frac{5+x}{x+35}\geq \frac{4}{9}$$

Denominator x+35 cannot be zero since division by zero is not defined. There are two cases.

$$x+35>0 x+35<0$$

Consider the case when x+35 is positive. Move 35 to the right hand side.

$$x>-35$$

The initial inequality does not change the direction when multiplied by x+35 for x+35>0.

$$5+x\geq \frac{4}{9}\left(x+35\right)$$

Multiply out the right hand side.

$$5+x\geq \frac{4}{9}x+\frac{140}{9}$$

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

$$x-\frac{4}{9}x\geq -5+\frac{140}{9}$$

Combine like terms.

$$\frac{5}{9}x\geq \frac{95}{9}$$

Divide both sides by \frac{5}{9}. Since \frac{5}{9} is positive, the inequality direction remains the same.

$$x\geq 19$$

Now consider the case when x+35 is negative. Move 35 to the right hand side.

$$x<-35$$

The initial inequality changes the direction when multiplied by x+35 for x+35<0.

$$5+x\leq \frac{4}{9}\left(x+35\right)$$

Multiply out the right hand side.

$$5+x\leq \frac{4}{9}x+\frac{140}{9}$$

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

$$x-\frac{4}{9}x\leq -5+\frac{140}{9}$$

Combine like terms.

$$\frac{5}{9}x\leq \frac{95}{9}$$

Divide both sides by \frac{5}{9}. Since \frac{5}{9} is positive, the inequality direction remains the same.

$$x\leq 19$$

Consider condition x<-35 specified above.

$$x<-35$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-35)\cup [19,\infty)$$