How do solve #(2x)/(x-2)<=3# algebraically?
2 Answers
Multiply both sides by
Distribute the 3
Move
Divide out negative one from both sides
*When dividing or multiplying by a negative, you have to switch the greater or less than sign.
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]}
