Solve 16/17=z/y+1/272y

$$\frac { 16 } { 17 } = \frac { z } { y } + \frac { 1 } { 272 y }$$

Solve for y

$$y=\frac{17z}{16}+\frac{1}{256}$$

$$z\neq -\frac{1}{272}$$

Steps for Solving Linear Equation

Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 272y, the least common multiple of 17,y,272y.

$$256y=272z+1$$

Divide both sides by 256.

$$\frac{256y}{256}=\frac{272z+1}{256}$$

Dividing by 256 undoes the multiplication by 256.

$$y=\frac{272z+1}{256}$$

Divide 272z+1 by 256.

$$y=\frac{17z}{16}+\frac{1}{256}$$

Variable y cannot be equal to 0.

$$y=\frac{17z}{16}+\frac{1}{256}\text{, }y\neq 0$$

Solve for z

$$z=\frac{16y}{17}-\frac{1}{272}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 272y, the least common multiple of 17,y,272y.

$$256y=272z+1$$

Swap sides so that all variable terms are on the left hand side.

$$272z+1=256y$$

Subtract 1 from both sides.

$$272z=256y-1$$

Divide both sides by 272.

$$\frac{272z}{272}=\frac{256y-1}{272}$$

Dividing by 272 undoes the multiplication by 272.

$$z=\frac{256y-1}{272}$$

Divide 256y-1 by 272.

$$z=\frac{16y}{17}-\frac{1}{272}$$