Solve 5(1)z=(n(y^2-2x+1)

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

Solve for n (complex solution)

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

Steps for Solving Linear Equation

Multiply 5 and 1 to get 5.

$$5z=n\left(y^{2}-2x+1\right)$$

Use the distributive property to multiply n by y^{2}-2x+1.

$$5z=ny^{2}-2nx+n$$

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

$$ny^{2}-2nx+n=5z$$

Combine all terms containing n.

$$\left(y^{2}-2x+1\right)n=5z$$

The equation is in standard form.

$$\left(1+y^{2}-2x\right)n=5z$$

Divide both sides by y^{2}-2x+1.

$$\frac{\left(1+y^{2}-2x\right)n}{1+y^{2}-2x}=\frac{5z}{1+y^{2}-2x}$$

Dividing by y^{2}-2x+1 undoes the multiplication by y^{2}-2x+1.

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

Solve for x (complex solution)

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

Steps for Solving Linear Equation

Multiply 5 and 1 to get 5.

$$5z=n\left(y^{2}-2x+1\right)$$

Use the distributive property to multiply n by y^{2}-2x+1.

$$5z=ny^{2}-2nx+n$$

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

$$ny^{2}-2nx+n=5z$$

Subtract ny^{2} from both sides.

$$-2nx+n=5z-ny^{2}$$

Subtract n from both sides.

$$-2nx=5z-ny^{2}-n$$

Reorder the terms.

$$-2nx=-ny^{2}+5z-n$$

The equation is in standard form.

$$\left(-2n\right)x=-ny^{2}+5z-n$$

Divide both sides by -2n.

$$\frac{\left(-2n\right)x}{-2n}=\frac{-ny^{2}+5z-n}{-2n}$$

Dividing by -2n undoes the multiplication by -2n.

$$x=\frac{-ny^{2}+5z-n}{-2n}$$

Divide -ny^{2}+5z-n by -2n.

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

Solve for n

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

Steps for Solving Linear Equation

Multiply 5 and 1 to get 5.

$$5z=n\left(y^{2}-2x+1\right)$$

Use the distributive property to multiply n by y^{2}-2x+1.

$$5z=ny^{2}-2nx+n$$

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

$$ny^{2}-2nx+n=5z$$

Combine all terms containing n.

$$\left(y^{2}-2x+1\right)n=5z$$

The equation is in standard form.

$$\left(1+y^{2}-2x\right)n=5z$$

Divide both sides by y^{2}-2x+1.

$$\frac{\left(1+y^{2}-2x\right)n}{1+y^{2}-2x}=\frac{5z}{1+y^{2}-2x}$$

Dividing by y^{2}-2x+1 undoes the multiplication by y^{2}-2x+1.

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

Solve for x

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

Steps for Solving Linear Equation

Multiply 5 and 1 to get 5.

$$5z=n\left(y^{2}-2x+1\right)$$

Use the distributive property to multiply n by y^{2}-2x+1.

$$5z=ny^{2}-2nx+n$$

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

$$ny^{2}-2nx+n=5z$$

Subtract ny^{2} from both sides.

$$-2nx+n=5z-ny^{2}$$

Subtract n from both sides.

$$-2nx=5z-ny^{2}-n$$

Reorder the terms.

$$-2nx=-ny^{2}+5z-n$$

The equation is in standard form.

$$\left(-2n\right)x=-ny^{2}+5z-n$$

Divide both sides by -2n.

$$\frac{\left(-2n\right)x}{-2n}=\frac{-ny^{2}+5z-n}{-2n}$$

Dividing by -2n undoes the multiplication by -2n.

$$x=\frac{-ny^{2}+5z-n}{-2n}$$

Divide -ny^{2}+5z-n by -2n.

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