What is (2n+3}{9(n-2)}-\frac{3n)/2 ?

The solution to (2n+3}{9(n-2)}-\frac{3n)/2 is (-27n^2+58n+6)/(18(n-2))

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simplify (2n+3}{9(n-2)}-\frac{3n)/2

simplify

\frac{2 n + 3}{9 ( n - 2 )} - \frac{3 n}{2}

\frac{- 27 n ^{2} + 58 n + 6}{18 ( n - 2 )}

Solution steps

\frac{2 n + 3}{9 ( n - 2 )} - \frac{3 n}{2}

Least Common Multiplier of

9 (n - 2), 2 : 18 (n - 2)

Adjust Fractions based on the LCM

= \frac{( 2 n + 3 ) \cdot 2}{18 ( n - 2 )} - \frac{27 n ( n - 2 )}{18 ( n - 2 )}

Apply the fraction rule:

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

= \frac{( 2 n + 3 ) \cdot 2 - 27 n ( n - 2 )}{18 ( n - 2 )}

Simplify

(2 n + 3) \cdot 2 - 27 n (n - 2) : - 27 n^{2} + 58 n + 6

= \frac{- 27 n ^{2} + 58 n + 6}{18 ( n - 2 )}

- 2 x + 4 = 0

2^{2} \cdot 2^{4}

f (n) = \frac{5 n ^{2} - 245}{7 n ^{2} + 112 n + 441}

f^{2} - 21 f = 0

(6.626 \cdot 1 0^{- 34}) (3 \cdot 1 0^{8})

What is (2n+3}{9(n-2)}-\frac{3n)/2 ?
The solution to (2n+3}{9(n-2)}-\frac{3n)/2 is (-27n^2+58n+6)/(18(n-2))