Solve 2x-9/4x+8-17leq0

$$\frac { 2 x - 9 } { 4 x + 8 } - 17 \leq 0$$

Solve for x

$$x\in (-\infty,-\frac{145}{66}]\cup (-2,\infty)$$

Solution Steps

Factor 4x+8.

$$\frac{2x-9}{4\left(x+2\right)}-17\leq 0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 17 times \frac{4\left(x+2\right)}{4\left(x+2\right)}.

$$\frac{2x-9}{4\left(x+2\right)}-\frac{17\times 4\left(x+2\right)}{4\left(x+2\right)}\leq 0$$

Since \frac{2x-9}{4\left(x+2\right)} and \frac{17\times 4\left(x+2\right)}{4\left(x+2\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2x-9-17\times 4\left(x+2\right)}{4\left(x+2\right)}\leq 0$$

Do the multiplications in 2x-9-17\times 4\left(x+2\right).

$$\frac{2x-9-68x-136}{4\left(x+2\right)}\leq 0$$

Combine like terms in 2x-9-68x-136.

$$\frac{-66x-145}{4\left(x+2\right)}\leq 0$$

Use the distributive property to multiply 4 by x+2.

$$\frac{-66x-145}{4x+8}\leq 0$$

For the quotient to be ≤0, one of the values -66x-145 and 4x+8 has to be ≥0, the other has to be ≤0, and 4x+8 cannot be zero. Consider the case when -66x-145\geq 0 and 4x+8 is negative.

$$-66x-145\geq 0 4x+8<0$$

The solution satisfying both inequalities is x\leq -\frac{145}{66}.

$$x\leq -\frac{145}{66}$$

Consider the case when -66x-145\leq 0 and 4x+8 is positive.

$$-66x-145\leq 0 4x+8>0$$

The solution satisfying both inequalities is x>-2.

$$x>-2$$

The final solution is the union of the obtained solutions.

$$x\leq -\frac{145}{66}\text{; }x>-2$$