What is the answer to 6z-4/(1+2z^-)=3 ?

The answer to 6z-4/(1+2z^-)=3 is z= 7/12 ;\quad 1+2z^-\ne 0

Solve for z: 6z-4/(1+2z^-)=3

The solution is z= 7/12 ;\quad 1+2z^-\ne 0

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6 z - \frac{4}{1 + 2 z ^{-}} = 3

z = \frac{7}{12}; 1 + 2 z^{-} \neq = 0

Decimal Notation

z = 0.58333 \dots

Solution steps

6 z - \frac{4}{1 + 2 z ^{-}} = 3

Multiply both sides by

1 + 2 z^{-}

6 z (1 + 2 z^{-}) - 4 = 3 (1 + 2 z^{-}); 1 + 2 z^{-} \neq = 0

Expand

6 z (1 + 2 z^{-}) - 4 : 18 z - 4

Expand

3 (1 + 2 z^{-}) : 3 + 6 z^{-}

18 z - 4 = 3 + 6 z^{-}

Move

4

to the right side

18 z = 6 z^{-} + 7

Move

6 z^{-}

to the left side

12 z = 7

Divide both sides by

12

z = \frac{7}{12}; 1 + 2 z^{-} \neq = 0

\frac{6}{2 x - 3} = \frac{4}{2 x + 5}

\frac{2}{x - 2} = 5 + \frac{x}{x - 2}

\frac{180 ( x - 2 )}{x} = 150

\frac{1}{x} - \frac{x}{9} = 0

\frac{10}{y} + \frac{14}{y} = 4

What is the answer to 6z-4/(1+2z^-)=3 ?
The answer to 6z-4/(1+2z^-)=3 is z= 7/12 ;\quad 1+2z^-\ne 0
Solve for z: 6z-4/(1+2z^-)=3
The solution is z= 7/12 ;\quad 1+2z^-\ne 0