Solve the Inequality for x 8/(x+2)>=1/(x+2)

$\frac{8}{x + 2} \geq \frac{1}{x + 2}$

Step 1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$8 \geq 1$

Step 2

Since $8 \geq 1$ , the inequality will always be true.

Always true

Step 3

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$