Solve x^2+2x+5/x+4geqx-3

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

Solve for x

$$x\in (-\infty,-17]\cup (-4,\infty)$$

Solution Steps

Subtract x from both sides.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{x+4}{x+4}.

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

Since \frac{x^{2}+2x+5}{x+4} and \frac{x\left(x+4\right)}{x+4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in x^{2}+2x+5-x\left(x+4\right).

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

Combine like terms in x^{2}+2x+5-x^{2}-4x.

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

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

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

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

$$x>-4$$

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

$$-2x+5\geq -3\left(x+4\right)$$

Multiply out the right hand side.

$$-2x+5\geq -3x-12$$

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

$$-2x+3x\geq -5-12$$

Combine like terms.

$$x\geq -17$$

Consider condition x>-4 specified above.

$$x>-4$$

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

$$x<-4$$

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

$$-2x+5\leq -3\left(x+4\right)$$

Multiply out the right hand side.

$$-2x+5\leq -3x-12$$

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

$$-2x+3x\leq -5-12$$

Combine like terms.

$$x\leq -17$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-17]\cup (-4,\infty)$$