What is expand (3/2 ,-3)y(1/2 ,((-3))/4) ?

The solution to expand (3/2 ,-3)y(1/2 ,((-3))/4) is 3/2 (1/2*((-3))/4 ,)y,-3(1/2*((-3))/4 ,)y

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expand (3/2 ,-3)y(1/2 ,((-3))/4)

expand

(\frac{3}{2}, - 3) y (\frac{1}{2}, \frac{( - 3 )}{4})

\frac{3}{2} (\frac{1}{2} \cdot \frac{( - 3 )}{4},) y, - 3 (\frac{1}{2} \cdot \frac{( - 3 )}{4},) y

Solution steps

(\frac{3}{2}, - 3) y (\frac{1}{2}, \frac{( - 3 )}{4})

= (\frac{3}{2}, - 3) (\frac{1}{2} \cdot \frac{( - 3 )}{4},) y

Apply the distributive law:

a (b - c) = ab - a c

a = y (\frac{1}{2}, \frac{( - 3 )}{4}), b = \frac{3}{2},, c = 3

= y (\frac{1}{2}, \frac{( - 3 )}{4}) \frac{3}{2}, - y (\frac{1}{2}, \frac{( - 3 )}{4}) \cdot 3

= \frac{3}{2} (\frac{1}{2} \cdot \frac{( - 3 )}{4},) y, - 3 (\frac{1}{2} \cdot \frac{( - 3 )}{4},) y

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

\frac{( - 1 ) ^{k + 1} k ( k + 1 )}{2}

e^{x} (e^{y} + e^{- x})

ln (x) (x + 1) (x + 2)

\frac{3}{( 2 + j w ) ( 5 + j w )}

What is expand (3/2 ,-3)y(1/2 ,((-3))/4) ?
The solution to expand (3/2 ,-3)y(1/2 ,((-3))/4) is 3/2 (1/2*((-3))/4 ,)y,-3(1/2*((-3))/4 ,)y