Solve (xquady-zquadz+3)+(yquad4quad3)=(4quad8quad16)

$$( x \quad y - z \quad z + 3 ) + ( y \quad 4 \quad 3 ) = ( 4 \quad 8 \quad 16 )$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{3y_{4}-z^{2}-509}{y}\text{, }&y\neq 0\\x\in \mathrm{C}\text{, }&y_{4}=\frac{z^{2}+509}{3}\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$xy-z^{2}+3+y_{4}\times 3=4\times 8\times 16$$

Multiply 4 and 8 to get 32.

$$xy-z^{2}+3+y_{4}\times 3=32\times 16$$

Multiply 32 and 16 to get 512.

$$xy-z^{2}+3+y_{4}\times 3=512$$

Add z^{2} to both sides.

$$xy+3+y_{4}\times 3=512+z^{2}$$

Subtract 3 from both sides.

$$xy+y_{4}\times 3=512+z^{2}-3$$

Subtract 3 from 512 to get 509.

$$xy+y_{4}\times 3=509+z^{2}$$

Subtract y_{4}\times 3 from both sides.

$$xy=509+z^{2}-y_{4}\times 3$$

Multiply -1 and 3 to get -3.

$$xy=509+z^{2}-3y_{4}$$

The equation is in standard form.

$$yx=509+z^{2}-3y_{4}$$

Divide both sides by y.

$$\frac{yx}{y}=\frac{509+z^{2}-3y_{4}}{y}$$

Dividing by y undoes the multiplication by y.

$$x=\frac{509+z^{2}-3y_{4}}{y}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{3y_{4}-z^{2}-509}{x}\text{, }&x\neq 0\\y\in \mathrm{C}\text{, }&y_{4}=\frac{z^{2}+509}{3}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$xy-z^{2}+3+y_{4}\times 3=4\times 8\times 16$$

Multiply 4 and 8 to get 32.

$$xy-z^{2}+3+y_{4}\times 3=32\times 16$$

Multiply 32 and 16 to get 512.

$$xy-z^{2}+3+y_{4}\times 3=512$$

Add z^{2} to both sides.

$$xy+3+y_{4}\times 3=512+z^{2}$$

Subtract 3 from both sides.

$$xy+y_{4}\times 3=512+z^{2}-3$$

Subtract 3 from 512 to get 509.

$$xy+y_{4}\times 3=509+z^{2}$$

Subtract y_{4}\times 3 from both sides.

$$xy=509+z^{2}-y_{4}\times 3$$

Multiply -1 and 3 to get -3.

$$xy=509+z^{2}-3y_{4}$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{509+z^{2}-3y_{4}}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{509+z^{2}-3y_{4}}{x}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{3y_{4}-z^{2}-509}{y}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&y_{4}=\frac{z^{2}+509}{3}\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$xy-z^{2}+3+y_{4}\times 3=4\times 8\times 16$$

Multiply 4 and 8 to get 32.

$$xy-z^{2}+3+y_{4}\times 3=32\times 16$$

Multiply 32 and 16 to get 512.

$$xy-z^{2}+3+y_{4}\times 3=512$$

Add z^{2} to both sides.

$$xy+3+y_{4}\times 3=512+z^{2}$$

Subtract 3 from both sides.

$$xy+y_{4}\times 3=512+z^{2}-3$$

Subtract 3 from 512 to get 509.

$$xy+y_{4}\times 3=509+z^{2}$$

Subtract y_{4}\times 3 from both sides.

$$xy=509+z^{2}-y_{4}\times 3$$

Multiply -1 and 3 to get -3.

$$xy=509+z^{2}-3y_{4}$$

The equation is in standard form.

$$yx=509+z^{2}-3y_{4}$$

Divide both sides by y.

$$\frac{yx}{y}=\frac{509+z^{2}-3y_{4}}{y}$$

Dividing by y undoes the multiplication by y.

$$x=\frac{509+z^{2}-3y_{4}}{y}$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{3y_{4}-z^{2}-509}{x}\text{, }&x\neq 0\\y\in \mathrm{R}\text{, }&y_{4}=\frac{z^{2}+509}{3}\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$xy-z^{2}+3+y_{4}\times 3=4\times 8\times 16$$

Multiply 4 and 8 to get 32.

$$xy-z^{2}+3+y_{4}\times 3=32\times 16$$

Multiply 32 and 16 to get 512.

$$xy-z^{2}+3+y_{4}\times 3=512$$

Add z^{2} to both sides.

$$xy+3+y_{4}\times 3=512+z^{2}$$

Subtract 3 from both sides.

$$xy+y_{4}\times 3=512+z^{2}-3$$

Subtract 3 from 512 to get 509.

$$xy+y_{4}\times 3=509+z^{2}$$

Subtract y_{4}\times 3 from both sides.

$$xy=509+z^{2}-y_{4}\times 3$$

Multiply -1 and 3 to get -3.

$$xy=509+z^{2}-3y_{4}$$

Divide both sides by x.

$$\frac{xy}{x}=\frac{509+z^{2}-3y_{4}}{x}$$

Dividing by x undoes the multiplication by x.

$$y=\frac{509+z^{2}-3y_{4}}{x}$$