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

1 Answer
Sep 22, 2016

Answer:

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

Explanation:

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

Split into cases #x < -2# and #x > -2#...

#color(white)()#
Case #bb(x < -2)#

If #x < -2# then #x + 2 < 0#

Multiply both sides of the inequality by #(x+2)# and reverse the inequality fo find:

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

Subtract #8# from both ends to get:

#0 >= 4x#

Divide both sides by #4# and transpose to find:

#x <= 0#

If #x < -2# then we already have #x <= 0#, so the whole of #(-oo, -2)# is part of the solution set.

#color(white)()#
Case #bb(x > -2)#

If #x > -2# then #x + 2 > 0#

Multiply both sides of the inequality by #(x+2)# to find:

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

Subtract #8# from both ends to get:

#0 <= 4x#

Divide both sides by #4# and transpose to find:

#x >= 0#

If #x >= 0# then #x > -2#, so the whole of #[0, oo)# is part of the solution set.

#color(white)()#
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]}