Solve 2x^2-5/x+1leq2x+7

$$\frac { 2 x ^ { 2 } - 5 } { x + 1 } \leq 2 x + 7$$

Solve for x

$$x\in (-\infty,-\frac{4}{3}]\cup (-1,\infty)$$

Solution Steps

Subtract 2x from both sides.

$$\frac{2x^{2}-5}{x+1}-2x\leq 7$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x times \frac{x+1}{x+1}.

$$\frac{2x^{2}-5}{x+1}+\frac{-2x\left(x+1\right)}{x+1}\leq 7$$

Since \frac{2x^{2}-5}{x+1} and \frac{-2x\left(x+1\right)}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{2x^{2}-5-2x\left(x+1\right)}{x+1}\leq 7$$

Do the multiplications in 2x^{2}-5-2x\left(x+1\right).

$$\frac{2x^{2}-5-2x^{2}-2x}{x+1}\leq 7$$

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

$$\frac{-5-2x}{x+1}\leq 7$$

Denominator x+1 cannot be zero since division by zero is not defined. There are two cases.

$$x+1>0 x+1<0$$

Consider the case when x+1 is positive. Move 1 to the right hand side.

$$x>-1$$

The initial inequality does not change the direction when multiplied by x+1 for x+1>0.

$$-5-2x\leq 7\left(x+1\right)$$

Multiply out the right hand side.

$$-5-2x\leq 7x+7$$

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

$$-2x-7x\leq 5+7$$

Combine like terms.

$$-9x\leq 12$$

Divide both sides by -9. Since -9 is negative, the inequality direction is changed.

$$x\geq -\frac{4}{3}$$

Consider condition x>-1 specified above.

$$x>-1$$

Now consider the case when x+1 is negative. Move 1 to the right hand side.

$$x<-1$$

The initial inequality changes the direction when multiplied by x+1 for x+1<0.

$$-5-2x\geq 7\left(x+1\right)$$

Multiply out the right hand side.

$$-5-2x\geq 7x+7$$

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

$$-2x-7x\geq 5+7$$

Combine like terms.

$$-9x\geq 12$$

Divide both sides by -9. Since -9 is negative, the inequality direction is changed.

$$x\leq -\frac{4}{3}$$

The final solution is the union of the obtained solutions.

$$x\in (-\infty,-\frac{4}{3}]\cup (-1,\infty)$$