What is expand 4(1-x)^{-1}-3(3+2x)^{-1} ?

The solution to expand 4(1-x)^{-1}-3(3+2x)^{-1} is (11x+9)/(-2x^2-x+3)

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expand 4(1-x)^{-1}-3(3+2x)^{-1}

expand

4 (1 - x)^{- 1} - 3 (3 + 2 x)^{- 1}

\frac{11 x + 9}{- 2 x ^{2} - x + 3}

Solution steps

4 (1 - x)^{- 1} - 3 (3 + 2 x)^{- 1}

4 (1 - x)^{- 1} = \frac{4}{1 - x}

3 (3 + 2 x)^{- 1} = \frac{3}{3 + 2 x}

= \frac{4}{- x + 1} - \frac{3}{2 x + 3}

Least Common Multiplier of

1 - x, 3 + 2 x : (- x + 1) (2 x + 3)

Adjust Fractions based on the LCM

= \frac{4 ( 2 x + 3 )}{( 1 - x ) ( 2 x + 3 )} - \frac{3 ( - x + 1 )}{( 3 + 2 x ) ( - x + 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{4 ( 2 x + 3 ) - 3 ( - x + 1 )}{( - x + 1 ) ( 2 x + 3 )}

Expand

(- x + 1) (2 x + 3) : - 2 x^{2} - x + 3

= \frac{4 ( 2 x + 3 ) - 3 ( - x + 1 )}{- 2 x ^{2} - x + 3}

Expand

4 (2 x + 3) - 3 (- x + 1) : 11 x + 9

= \frac{11 x + 9}{- 2 x ^{2} - x + 3}

\frac{1}{28} (2 ln (2) + 1)

\frac{1}{( 2 x + y + 1 ) ^{2}}

6 (2 + 3 lo g_{2} (5))

2 ((3 x + 4 y) (4 y - 3 x) (3 x - 4 y))

144 (\frac{7 y}{9} + \frac{3}{4})

What is expand 4(1-x)^{-1}-3(3+2x)^{-1} ?
The solution to expand 4(1-x)^{-1}-3(3+2x)^{-1} is (11x+9)/(-2x^2-x+3)