What is expand 1/3 (2+1/5 (4z-1)= 3/4 z-1/3) ?

The solution to expand 1/3 (2+1/5 (4z-1)= 3/4 z-1/3) is (4z}{15}-1/15+2/3 =\frac{3z)/4-1/9

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expand 1/3 (2+1/5 (4z-1)= 3/4 z-1/3)

expand

\frac{1}{3} (2 + \frac{1}{5} (4 z - 1) = \frac{3}{4} z - \frac{1}{3})

\frac{4 z}{15} - \frac{1}{15} + \frac{2}{3} = \frac{3 z}{4} - \frac{1}{9}

Solution steps

\frac{1}{3} (2 + \frac{1}{5} (4 z - 1) = \frac{3}{4} z - \frac{1}{3})

= \frac{1}{3} (2 + \frac{1}{5} \cdot \frac{3}{4} = z (4 z - 1) - \frac{1}{3})

Distribute parentheses

\frac{1}{3} \cdot 2 + \frac{1}{3} \cdot \frac{1}{5} (4 z - 1) = \frac{3}{4} z + \frac{1}{3} (- \frac{1}{3})

Apply minus-plus rules

+ (- a) = - a

2 \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{5} (4 z - 1) = \frac{3}{4} z - \frac{1}{3} \cdot \frac{1}{3}

Simplify

2 \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{1}{5} (4 z - 1) = \frac{3}{4} z - \frac{1}{3} \cdot \frac{1}{3} : \frac{4 z}{15} - \frac{1}{15} + \frac{2}{3} = \frac{3 z}{4} - \frac{1}{9}

\frac{4 z}{15} - \frac{1}{15} + \frac{2}{3} = \frac{3 z}{4} - \frac{1}{9}

2 e^{- x^{2} - y^{2}} y - 2 e^{- x^{2} - y^{2}} y (y^{2} - x^{2})

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

x H (x - 1) + (4 - 2 x) H (x - 2) - (3 - x) H (x - 3)

(3 cos (x) - 7 sin (x)) d x

\frac{( x ^{2} + 4 x ) ^{2}}{e ^{2 x}}

What is expand 1/3 (2+1/5 (4z-1)= 3/4 z-1/3) ?
The solution to expand 1/3 (2+1/5 (4z-1)= 3/4 z-1/3) is (4z}{15}-1/15+2/3 =\frac{3z)/4-1/9