How do you solve #(-x+8)/(x-2)>=5#?

1 Answer
Dec 24, 2016

Answer:

#x<=3#

Explanation:

If we start with #(-x+8)/(x-2)>=5#, then our first step should be to get rid of the denominator. To do that, we multiply both sides by #(x-2)#. That changes the eqaution into #(-x+8)/cancel(x-2)>=5*(x-2)#. If we distribute the #5#, it becomes #-x+8>=5x-10#. Just add #x# and #10# on both sides, and we are left with #18>=6x#. Divide by #6# on each side, and we see that #3>=x#. The conventional way to write inequalities is to have the variable first, so we shouls rewrite it as #x<=3#