What is expand 6((x+1)/8-(2x-3)/(16)) ?

The solution to expand 6((x+1)/8-(2x-3)/(16)) is (3(x+1))/4-(3(2x-3))/8

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expand 6((x+1)/8-(2x-3)/(16))

expand

6 (\frac{x + 1}{8} - \frac{2 x - 3}{16})

\frac{3 ( x + 1 )}{4} - \frac{3 ( 2 x - 3 )}{8}

Solution steps

6 (\frac{x + 1}{8} - \frac{2 x - 3}{16})

Apply the distributive law:

a (b - c) = ab - a c

6 (\frac{x + 1}{8} - \frac{2 x - 3}{16}) = 6 \cdot \frac{x + 1}{8} - 6 \cdot \frac{2 x - 3}{16}

= 6 \cdot \frac{x + 1}{8} - 6 \cdot \frac{2 x - 3}{16}

Simplify

6 \cdot \frac{x + 1}{8} - 6 \cdot \frac{2 x - 3}{16} : \frac{3 ( x + 1 )}{4} - \frac{3 ( 2 x - 3 )}{8}

= \frac{3 ( x + 1 )}{4} - \frac{3 ( 2 x - 3 )}{8}

(x^{2} + 3 x + 1 \circ 3 x + 5) (3)

2 x (cos (y) - y sin (y))

- \frac{ln ( 2 )}{45 ( 1 - e ^{\frac{l n ( 2 )}{45}} )}

\frac{4 x}{( x + 2 ) ( x - 2 )}

\frac{d ^{4}}{d x ^{4}} ((y) (2 x^{5} - 2 x))

What is expand 6((x+1)/8-(2x-3)/(16)) ?
The solution to expand 6((x+1)/8-(2x-3)/(16)) is (3(x+1))/4-(3(2x-3))/8