How do you simplify and find the excluded value of $\frac{6 - y}{y^{2} - 2 y - 24}$ $(6-y)/(y^2-2y-24)$ ?

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Answer 1 · 2017-06-13T03:50:38

factor the denominator then reduce

$y^{2} - 2 y - 24$ $y^2-2y-24$ factors to $(y + 4) (y - 6)$ $(y+4)(y-6)$ .
Now factor a -1 out of the numerator $6 - y = - 1 (y - 6)$ $6-y = -1(y-6)$

The factored problem is now $\frac{- 1 (y - 6)}{(y + 4) (y - 6)}$ $[-1(y-6)]/[(y+4)(y-6)$

reduce to get $- \frac{1}{y + 4}$ $-1/(y+4)$

Answer 2 · 2017-06-13T03:52:48

$= - \frac{1}{(y + 4)}$ $= -1/((y + 4))$

excluded value when $y = - 4$ $y = -4$

$\frac{6 - y}{y^{2} - 2 y - 24} = \frac{6 - y}{(y - 6) (y + 4)}$ $(6 - y)/(y^2 - 2 y - 24) = (6 - y)/((y - 6)(y + 4))$

$= (6 - y)/(-(-y + 6)(y + 4)) = cancel(6 - y)/(-cancel((6 - y))(y + 4))$

$= 1/(-(y + 4)) = -1/((y + 4))$

excluded value when $(y + 4) =0 -> y = -4$ therefore $y = -4$

How do you simplify and find the excluded value of (6-y)/(y^2-2y-24)6−yy2−2y−24(6-y)/(y^2-2y-24)?