How do solve #(2x)/(x-2)<=3# algebraically?

2 Answers
Aug 6, 2018

Multiply both sides by #x-2#

#2x<=3(x-2)#

Distribute the 3

#2x<=3x-6#

Move #x# terms to one side

#-x<=-6#

Divide out negative one from both sides

#x>=6#

*When dividing or multiplying by a negative, you have to switch the greater or less than sign.

Aug 7, 2018

Answer:

#x in(-00,2)uu[6,oo)#

Explanation:

#"subtract 3 from both sides"#

#(2x)/(x-2)-3<=0#

#"combine the left side as a single fraction"#

#(2x)/(x-2)-(3(x-2))/(x-2)<=0#

#(6-x)/(x-2)<=0#

#"find the critical values of numerator/denominator"#

#6-x=0rArrx=6larrcolor(blue)"is a zero"#

#x-2=0rArrx=2#

#"these values divide the domain into 3 intervals"#

#(-oo,2)uu(2,6]uu[6,oo)#

#"select a value for x as a "color(red)"test point in each interval"#

#x=1to(6-1)/(1-2)=-5<0larrcolor(blue)"valid"#

#x=3to(6-3)/(3-2)=3>0larrcolor(blue)"not valid"#

#x=10to(6-10)/(10-2)=-1/2<0larrcolor(blue)"valid"#

#x in(-oo,2)uu[6,oo)#
graph{(2x)/(x-2)-3 [-10, 10, -5, 5]}