Solve 17xy+1/x-3=z

$$17 x y + \frac { 1 } { x } - 3 = z$$

Solve for y

$$y=-\frac{1-3x-xz}{17x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$17xyx+1+x\left(-3\right)=zx$$

Multiply x and x to get x^{2}.

$$17x^{2}y+1+x\left(-3\right)=zx$$

Subtract 1 from both sides.

$$17x^{2}y+x\left(-3\right)=zx-1$$

Subtract x\left(-3\right) from both sides.

$$17x^{2}y=zx-1-x\left(-3\right)$$

Multiply -1 and -3 to get 3.

$$17x^{2}y=zx-1+3x$$

The equation is in standard form.

$$17x^{2}y=xz+3x-1$$

Divide both sides by 17x^{2}.

$$\frac{17x^{2}y}{17x^{2}}=\frac{xz+3x-1}{17x^{2}}$$

Dividing by 17x^{2} undoes the multiplication by 17x^{2}.

$$y=\frac{xz+3x-1}{17x^{2}}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{9+6z+z^{2}-68y}+z+3}{34y}\text{; }x=\frac{-\sqrt{9+6z+z^{2}-68y}+z+3}{34y}\text{, }&y\neq 0\text{ and }y\leq \frac{\left(z+3\right)^{2}}{68}\\x=\frac{1}{z+3}\text{, }&y=0\text{ and }z\neq -3\end{matrix}\right.$$