Solve (x-3)(x+6)-(x-8)/x+4geqx+1

$$\frac { ( x - 3 ) ( x + 6 ) - ( x - 8 ) } { x + 4 } \geq x + 1$$

Solve for x

$$x\in [-\frac{14}{3},-4)$$

Solution Steps

Use the distributive property to multiply x-3 by x+6 and combine like terms.

$$\frac{x^{2}+3x-18-\left(x-8\right)}{x+4}\geq x+1$$

To find the opposite of x-8, find the opposite of each term.

$$\frac{x^{2}+3x-18-x+8}{x+4}\geq x+1$$

Combine 3x and -x to get 2x.

$$\frac{x^{2}+2x-18+8}{x+4}\geq x+1$$

Add -18 and 8 to get -10.

$$\frac{x^{2}+2x-10}{x+4}\geq x+1$$

Subtract x from both sides.

$$\frac{x^{2}+2x-10}{x+4}-x\geq 1$$

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

$$\frac{x^{2}+2x-10}{x+4}-\frac{x\left(x+4\right)}{x+4}\geq 1$$

Since \frac{x^{2}+2x-10}{x+4} and \frac{x\left(x+4\right)}{x+4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{2}+2x-10-x\left(x+4\right)}{x+4}\geq 1$$

Do the multiplications in x^{2}+2x-10-x\left(x+4\right).

$$\frac{x^{2}+2x-10-x^{2}-4x}{x+4}\geq 1$$

Combine like terms in x^{2}+2x-10-x^{2}-4x.

$$\frac{-2x-10}{x+4}\geq 1$$

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.

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

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

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

Combine like terms.

$$-3x\geq 14$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\leq -\frac{14}{3}$$

Consider condition x>-4 specified above.

$$x\in \emptyset$$

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.

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

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

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

Combine like terms.

$$-3x\leq 14$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\geq -\frac{14}{3}$$

Consider condition x<-4 specified above.

$$x\in [-\frac{14}{3},-4)$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{14}{3},-4)$$