Solve 2x+3y+4z-6x*5y=42806

$$2x+3y+4z-6x \times 5y=42806$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{3y+4z-42806}{2\left(1-15y\right)}\text{, }&y\neq \frac{1}{15}\\x\in \mathrm{R}\text{, }&y=\frac{1}{15}\text{ and }z=\frac{214029}{20}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 6 and 5 to get 30.

$$2x+3y+4z-30xy=42806$$

Subtract 3y from both sides.

$$2x+4z-30xy=42806-3y$$

Subtract 4z from both sides.

$$2x-30xy=42806-3y-4z$$

Combine all terms containing x.

$$\left(2-30y\right)x=42806-3y-4z$$

The equation is in standard form.

$$\left(2-30y\right)x=42806-4z-3y$$

Divide both sides by 2-30y.

$$\frac{\left(2-30y\right)x}{2-30y}=\frac{42806-4z-3y}{2-30y}$$

Dividing by 2-30y undoes the multiplication by 2-30y.

$$x=\frac{42806-4z-3y}{2-30y}$$

Divide 42806-3y-4z by 2-30y.

$$x=\frac{42806-4z-3y}{2\left(1-15y\right)}$$

Solve for y

$$\left\{\begin{matrix}y=-\frac{2\left(x+2z-21403\right)}{3\left(1-10x\right)}\text{, }&x\neq \frac{1}{10}\\y\in \mathrm{R}\text{, }&x=\frac{1}{10}\text{ and }z=\frac{214029}{20}\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply 6 and 5 to get 30.

$$2x+3y+4z-30xy=42806$$

Subtract 2x from both sides.

$$3y+4z-30xy=42806-2x$$

Subtract 4z from both sides.

$$3y-30xy=42806-2x-4z$$

Combine all terms containing y.

$$\left(3-30x\right)y=42806-2x-4z$$

The equation is in standard form.

$$\left(3-30x\right)y=42806-4z-2x$$

Divide both sides by 3-30x.

$$\frac{\left(3-30x\right)y}{3-30x}=\frac{42806-4z-2x}{3-30x}$$

Dividing by 3-30x undoes the multiplication by 3-30x.

$$y=\frac{42806-4z-2x}{3-30x}$$

Divide 42806-2x-4z by 3-30x.

$$y=\frac{2\left(21403-2z-x\right)}{3\left(1-10x\right)}$$