Answer :

[tex]\dfrac{2y}{y^2-9y+18}-\dfrac{1}{y-6}=\\ \dfrac{2y}{y^2-3y-6y+18}-\dfrac{1}{y-6}=\\ \dfrac{2y}{y(y-3)-6(y-3)}-\dfrac{1}{y-6}=\\ \dfrac{2y}{(y-6)(y-3)}-\dfrac{1}{y-6}=\\ \dfrac{2y}{(y-6)(y-3)}-\dfrac{y-3}{(y-6)(y-3)}=\\ \dfrac{2y-y+3}{(y-6)(y-3)}=\\ \dfrac{y+3}{(y-6)(y-3)}[/tex]

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