How do solve (2x)/(x-2)<=32xx−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"subtract 3 from both sides
(2x)/(x-2)-3<=02xx−2−3≤0
"combine the left side as a single fraction"combine the left side as a single fraction
(2x)/(x-2)-(3(x-2))/(x-2)<=02xx−2−3(x−2)x−2≤0
(6-x)/(x-2)<=06−xx−2≤0
"find the critical values of numerator/denominator"find the critical values of numerator/denominator
6-x=0rArrx=6larrcolor(blue)"is a zero"6−x=0⇒x=6←is a zero
x-2=0rArrx=2x−2=0⇒x=2
"these values divide the domain into 3 intervals"these values divide the domain into 3 intervals
(-oo,2)uu(2,6]uu[6,oo)(−∞,2)∪(2,6]∪[6,∞)
"select a value for x as a "color(red)"test point in each interval"select a value for x as a test point in each interval
x=1to(6-1)/(1-2)=-5<0larrcolor(blue)"valid"x=1→6−11−2=−5<0←valid
x=3to(6-3)/(3-2)=3>0larrcolor(blue)"not valid"x=3→6−33−2=3>0←not valid
x=10to(6-10)/(10-2)=-1/2<0larrcolor(blue)"valid"x=10→6−1010−2=−12<0←valid
x in(-oo,2)uu[6,oo)x∈(−∞,2)∪[6,∞)
graph{(2x)/(x-2)-3 [-10, 10, -5, 5]}
