What is (-n^2+16n-63)/(n^2-2n-35) ?

The solution to (-n^2+16n-63)/(n^2-2n-35) is -(n-9)/(n+5)

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simplify (-n^2+16n-63)/(n^2-2n-35)

simplify

\frac{- n ^{2} + 16 n - 63}{n ^{2} - 2 n - 35}

- \frac{n - 9}{n + 5}

Solution steps

\frac{- n ^{2} + 16 n - 63}{n ^{2} - 2 n - 35}

Factor

- n^{2} + 16 n - 63 : - (n - 7) (n - 9)

= \frac{- ( n - 7 ) ( n - 9 )}{n ^{2} - 2 n - 35}

Factor

n^{2} - 2 n - 35 : (n + 5) (n - 7)

= \frac{- ( n - 7 ) ( n - 9 )}{( n + 5 ) ( n - 7 )}

Cancel the common factor:

n - 7

= \frac{- ( n - 9 )}{n + 5}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{n - 9}{n + 5}

∣ 3 x - (4 x - 2) ∣ = 4

48 = 40 + k \cdot 4

8 w + 11 - 9 w < - 2 w - 5

x^{2} - 3 x + 2 = - 1

11 (k - 1) = 7 (k - 9)

What is (-n^2+16n-63)/(n^2-2n-35) ?
The solution to (-n^2+16n-63)/(n^2-2n-35) is -(n-9)/(n+5)