# Solve this inequation

• Feb 23rd 2013, 07:02 PM
Razbaz
Solve this inequation
(y^2 - 6)/y <= 1
Any help would be appreciated.
• Feb 23rd 2013, 09:08 PM
Soroban
Re: Solve this inequation
Hello, Razbaz!

Quote:

$\text{Solve: }\:\frac{y^2 - 6}{y} \:\le \:1$

$\text{We have: }\:\frac{y^2-6}{y} - 1 \:\le\:0 \quad\Rightarrow\quad \frac{y^2-6-y}{y} \:\le\:1$

. . . . . . . . . $\frac{y^2-y-6}{y}\:\le\:0 \quad\Rightarrow\quad \frac{(y-3)(y+2)}{y}\:\le\:0$

We have three critical values: . $y \;=\;\text{-}2,\,0,\,3$
. . which divides the number line into four intervals.

. . $\begin{array}{ccccccc}---&\bullet & --- & \bullet & --- & \bullet & --- \\ & \text{-}2 && 0 && 3 \end{array}$

Test a value of $y$ in each interval.

$\text{On }(\text{-}\infty,\,\text{-}2), \text{ let }y = \text{-}3\!:\;\frac{(\text{-}3)^2-6}{\text{-}3} \:=\:\frac{3}{\text{-}3} \:=\:\text{-}1\;\;\text{ Yes}$

$\text{On }(\text{-}2,\,0),\text{ let }y = \text{-}1\!:\:\frac{(\text{-}1)^2-6}{\text{-}1} \:=\:\frac{\text{-}5}{\text{-}1} \:=\:5 \;\;\text{ no}$

$\text{On }(0,\,3),\text{ let }y = 1\!:\;\frac{1^2-6}{1} \:=\:-5\;\;\text{ Yes}$

$\text{On }(3,\,\infty),\text{ let }y = 4\!:\:\frac{4^2-6}{4} \:=\:\frac{10}{4}\;\;\text{ no}$

$\text{Therefore: }\:(\text{-}\infty,\,2]\:\cup\:(0,\,3]$