Question

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

Answer 1

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)`