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\[\frac{ b+2 }{ 2b^2-b-1 } - \frac{ 1 }{ b-1 }\] A. −1/2b+1 B. −b+1/(2b+1)(b−1) C. b+1/(2b+1)(b−1) D. b+1/2b^2
Facor 2b^2-b-1 first
How?
(2b+1)(b-1) = (2b*b)-2b+b-1 = 2b^2-b-1
yup
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\[\frac{ b+2 }{ (2b+1)(b-1) } - \frac{ 1 }{ (b-1) }\] This is the new eqation
\(\Large \frac{ b+2 }{ 2b^2-b-1 } - \frac{ 1 }{ b-1 }\) \(\Large =\frac{ b+2 }{ (2b+1)(b-1) } - \frac{ 1 }{ (b-1) }\) \(\Large =\frac{ b+2 }{ (2b+1)(b-1) } - \frac{2b+1}{(2b+1)(b-1)}\) \(\Large =\frac{b+2-2b-1}{(2b+1)(b-1)}\) \(\Large =\frac{-(b-1)}{(2b+1)(b-1)}=\color{blue}{\frac{-1}{2b+1} }\)
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