What is the answer to solvefor x,((x+2y))/(0.5)z=128 ?

The answer to solvefor x,((x+2y))/(0.5)z=128 is x=(64)/z-2y;\quad z\ne 0

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solvefor x,((x+2y))/(0.5)z=128

solve for

x, \frac{( x + 2 y )}{0.5} z = 128

x = \frac{64}{z} - 2 y; z \neq = 0

Solution steps

\frac{( x + 2 y )}{0.5} z = 128

Multiply fractions:

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

\frac{z ( x + 2 y )}{0.5} = 128

Multiply both sides by

0.5

z (x + 2 y) = 64

Divide both sides by

z; z \neq = 0

x + 2 y = \frac{64}{z}; z \neq = 0

Move

2 y

to the right side

x = \frac{64}{z} - 2 y; z \neq = 0

2 \frac{x}{3} = - 10

\frac{3}{4} x - 20 = \frac{1}{4} x - 10

x, - x + 3 y - z = - 3

f^{1} = 25 n, f^{2} = 25 n f^{3} = 15 n f^{4} = 12

\frac{17}{3} = \frac{x}{1.5}

What is the answer to solvefor x,((x+2y))/(0.5)z=128 ?
The answer to solvefor x,((x+2y))/(0.5)z=128 is x=(64)/z-2y;\quad z\ne 0