Solve L1:x-2y-4=0

$$L 1 : x - 2 y - 4 = 0$$

Solve for L_1

$$L_{1}=2x\left(y+2\right)$$

$$x\neq 0$$

Solution Steps

Multiply both sides of the equation by x.

$$L_{1}-2yx+x\left(-4\right)=0$$

Add 2yx to both sides. Anything plus zero gives itself.

$$L_{1}+x\left(-4\right)=2yx$$

Subtract x\left(-4\right) from both sides.

$$L_{1}=2yx-x\left(-4\right)$$

Multiply -1 and -4 to get 4.

$$L_{1}=2yx+4x$$

Solve for x

$$\left\{\begin{matrix}x=\frac{L_{1}}{2\left(y+2\right)}\text{, }&L_{1}\neq 0\text{ and }y\neq -2\\x\neq 0\text{, }&y=-2\text{ and }L_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.

$$L_{1}-2yx+x\left(-4\right)=0$$

Subtract L_{1} from both sides. Anything subtracted from zero gives its negation.

$$-2yx+x\left(-4\right)=-L_{1}$$

Combine all terms containing x.

$$\left(-2y-4\right)x=-L_{1}$$

Divide both sides by -2y-4.

$$\frac{\left(-2y-4\right)x}{-2y-4}=-\frac{L_{1}}{-2y-4}$$

Dividing by -2y-4 undoes the multiplication by -2y-4.

$$x=-\frac{L_{1}}{-2y-4}$$

Divide -L_{1} by -2y-4.

$$x=\frac{L_{1}}{2\left(y+2\right)}$$

Variable x cannot be equal to 0.

$$x=\frac{L_{1}}{2\left(y+2\right)}\text{, }x\neq 0$$