How do solve #3/(x-3)<=2/(x+2)# and write the answer as a inequality and interval notation?

1 Answer
Jun 13, 2017

Answer:

Inequality: #x <= -12# and #-2 < x < 3#
Interval: #(-infty,-12]# and #(-2,3)#

Explanation:

Step 1. Find the critical values by assuming equality.

Assume #3/(x-3) = 2/(x+2)#

You have critical values at #x=3# and #x=-2# because these would cause the equation to divide by zero.

Also, solving for #x# gives the last critical value

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

#3x+6 = 2x-6#

#x = -12#

Step 2. Evaluate the inequality around these critical values.

#{:("Crit. Value ","Test value ", 3/(x-3) <= 2/(x+2)),(x <= -12 ," "-20," True"),(-12 < x < -2, " "-10, " False"),(-2 < x < 3," "0," True"),(x > 3, " "5," False"):}#

Step 3. Complete by writing the inequality and interval notation.

Inequality: #" "x <= -12# and #-2 < x < 3#
Interval: #" "(-infty,-12]# and #(-2,3)#