Question

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

Answer 1

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.

Answer 2

`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]}