How do you solve #\frac { x - 8} { x - 4} > 0#?

2 Answers

Here manuplate th numerator to relate with denominator,
#(x-4-4)/(x-8)# ,
now seprating numerator
#(x-4)/(x-4)-(4)/(x-4)>0#
thus,
#1-4/(x-4)>0#
hence
#1>4/(x-4)#

#(x-4)>4#

#x>8#
to satisfy the above question

Jul 31, 2017

#(-oo,4)uu(8,+oo)#

Explanation:

#"the zeros of the numerator/denominator are"#

#"numerator "x=8," denominator "x=4#

#"these indicate where the rational function may change"#
#"sign"#

#"the intervals on the domain are"#

#x<4,color(white)(x)4 < x <8,color(white)(x)x>8#

#"consider a "color(blue)"test point " "in each interval"#

#"we want to find where the function is positive " >0#

#"substitute each test point into the function and consider"#
#"its sign"#

#color(magenta)"x = 3"to(-)/(-)tocolor(red)" positive"#

#color(magenta)"x=5"to(-)/(+)tocolor(blue)" negative"#

#color(magenta)"x = 10"to(+)/(+)tocolor(red)" positive"#

#rArr(-oo,4)uu(8,+oo)" is the solution"#
graph{(x-8)/(x-4) [-10, 10, -5, 5]}