# How do you solve #\frac{8}{x + 2} \leq 4#?

##### 1 Answer

#### Answer:

#### Explanation:

Given

Split into cases

**Case #bb(x < -2)#**

If

Multiply both sides of the inequality by **and reverse the inequality** fo find:

#8 >= 4(x+2) = 4x+8#

Subtract

#0 >= 4x#

Divide both sides by

#x <= 0#

If

**Case #bb(x > -2)#**

If

Multiply both sides of the inequality by

#8 <= 4(x+2) = 4x+8#

Subtract

#0 <= 4x#

Divide both sides by

#x >= 0#

If

**Conclusion**

Combining what we found from both cases we have:

#x in (-oo, -2) uu [0, oo)#

graph{(8/(x+2)-y)(y-4) = 0 [-20.13, 19.87, -8.64, 11.36]}