Solve z=2x^2y+5x-1

$$z = 2 { x }^{ 2 } y+5x-1$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{5x-z-1}{2x^{2}}\text{, }&x\neq 0\\y\in \mathrm{C}\text{, }&z=-1\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2x^{2}y+5x-1=z$$

Subtract 5x from both sides.

$$2x^{2}y-1=z-5x$$

Add 1 to both sides.

$$2x^{2}y=z-5x+1$$

The equation is in standard form.

$$2x^{2}y=1+z-5x$$

Divide both sides by 2x^{2}.

$$\frac{2x^{2}y}{2x^{2}}=\frac{1+z-5x}{2x^{2}}$$

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

$$y=\frac{1+z-5x}{2x^{2}}$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{5x-z-1}{2x^{2}}\text{, }&x\neq 0\\y\in \mathrm{R}\text{, }&z=-1\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$2x^{2}y+5x-1=z$$

Subtract 5x from both sides.

$$2x^{2}y-1=z-5x$$

Add 1 to both sides.

$$2x^{2}y=z-5x+1$$

The equation is in standard form.

$$2x^{2}y=1+z-5x$$

Divide both sides by 2x^{2}.

$$\frac{2x^{2}y}{2x^{2}}=\frac{1+z-5x}{2x^{2}}$$

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

$$y=\frac{1+z-5x}{2x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{8yz+8y+25}-5}{4y}\text{; }x=-\frac{\sqrt{8yz+8y+25}+5}{4y}\text{, }&y\neq 0\\x=\frac{z+1}{5}\text{, }&y=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{8yz+8y+25}-5}{4y}\text{; }x=-\frac{\sqrt{8yz+8y+25}+5}{4y}\text{, }&\left(y>0\text{ or }z\leq -\frac{8y+25}{8y}\right)\text{ and }\left(z\leq \text{Indeterminate}\text{ or }y\neq 0\right)\text{ and }\left(y<0\text{ or }\left(y\neq 0\text{ and }z\geq -\frac{8y+25}{8y}\right)\right)\\x=\frac{z+1}{5}\text{, }&y=0\end{matrix}\right.$$