How do you solve #\frac { 4} { x - 5} + 2\leq 4#?

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Answer 1 · 2018-06-26T20:08:50

#x\in(-\infty, 5)\cup(7, \infty)#

#\frac{4}{x-5}+2\le4#

#\frac{4}{x-5}+2-4\le0#

#\frac{4}{x-5}-2\le0#

#\frac{4-2x+10}{x-5}\le0#

#\frac{-2x+14}{x-5}\le0#

#\frac{-2(x-7)}{x-5}\le0#

#\frac{x-7}{x-5}\ge0#

Solving above inequality we get the solution

#x\in(-\infty, 5)\cup(7, \infty)#