Solve 2x+7/2(x+2)leq-1/x+2

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

Solve for x

$$x\in [-\frac{9}{2},-2)$$

Solution Steps

Use the distributive property to multiply 2 by x+2.

$$\frac{2x+7}{2x+4}\leq -\frac{1}{x+2}$$

Add \frac{1}{x+2} to both sides.

$$\frac{2x+7}{2x+4}+\frac{1}{x+2}\leq 0$$

Factor 2x+4.

$$\frac{2x+7}{2\left(x+2\right)}+\frac{1}{x+2}\leq 0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(x+2\right) and x+2 is 2\left(x+2\right). Multiply \frac{1}{x+2} times \frac{2}{2}.

$$\frac{2x+7}{2\left(x+2\right)}+\frac{2}{2\left(x+2\right)}\leq 0$$

Since \frac{2x+7}{2\left(x+2\right)} and \frac{2}{2\left(x+2\right)} have the same denominator, add them by adding their numerators.

$$\frac{2x+7+2}{2\left(x+2\right)}\leq 0$$

Combine like terms in 2x+7+2.

$$\frac{2x+9}{2\left(x+2\right)}\leq 0$$

Use the distributive property to multiply 2 by x+2.

$$\frac{2x+9}{2x+4}\leq 0$$

For the quotient to be ≤0, one of the values 2x+9 and 2x+4 has to be ≥0, the other has to be ≤0, and 2x+4 cannot be zero. Consider the case when 2x+9\geq 0 and 2x+4 is negative.

$$2x+9\geq 0 2x+4<0$$

The solution satisfying both inequalities is x\in \left[-\frac{9}{2},-2\right).

$$x\in [-\frac{9}{2},-2)$$

Consider the case when 2x+9\leq 0 and 2x+4 is positive.

$$2x+9\leq 0 2x+4>0$$

This is false for any x.

$$x\in \emptyset$$

The final solution is the union of the obtained solutions.

$$x\in [-\frac{9}{2},-2)$$