Given: The expression below
[tex]\frac{y^2-3y-18}{y^2-9y+18}[/tex]
To Determine: The lowest term of the given rational fraction
Solution
Let simplify both the numerator and the denominator
[tex]\begin{gathered} Numerator:y^2-3y-18 \\ y^2-3y-18=y^2-6y+3y-18 \\ y^2-3y-18=y(y-6)+3(y-6) \\ y^2-3y-18=(y-6)(y+3) \end{gathered}[/tex][tex]\begin{gathered} Denominator:y^2-9y+18 \\ y^2-9y+18=y^2-3y-6y+18 \\ y^2-9y+18=y(y-3)-6(y-3) \\ y^2-9y+18=(y-3)(y-6) \end{gathered}[/tex]
Therefore
[tex]\begin{gathered} \frac{y^2-3y-18}{y^2-9y+18}=\frac{(y-6)(y+3)}{(y-3)(y-6)} \\ y-6-is\text{ common} \\ \frac{y^{2}-3y-18}{y^{2}-9y+18}=\frac{(y-6)(y+3)}{(y-3)(y-6)} \\ \frac{y^{2}-3y-18}{y^{2}-9y+18}=\frac{y+3}{y-3} \end{gathered}[/tex]
Hence, the rational expression in its lowest term is
[tex]\frac{y+3}{y-3}[/tex]
The variable for the original expression is as given as
[tex]\begin{gathered} \frac{y^{2}-3y-18}{y^{2}-9y+18}=\frac{(y-6)(y+3)}{(y-3)(y-6)} \\ y\ne3,y\ne6 \end{gathered}[/tex]
