Solve {l}{quad(mathscr{C}):y=-2x+4/3x-3}{text{aupointdabscisse}-3}{text{(exemple}y=3x+2)}

$$\left. \begin{array} { l } { \quad ( \mathscr { C } ) : y = \frac { - 2 x + 4 } { 3 x - 3 } } \\ { \text { aupoint dabscisse } - 3 } \\ { \text { (exemple } y = 3 x + 2 ) } \end{array} \right.$$

Solve for x

$$x=\frac{4y+3C}{2y+3C}$$

$$y\neq 0\text{ and }C\neq -\frac{2y}{3}$$

Steps for Solving Linear Equation

Variable x cannot be equal to 1 since division by zero is not defined. Multiply both sides of the equation by 3y\left(x-1\right), the least common multiple of y,3x-3.

$$\left(3x-3\right)C=y\left(-2x+4\right)$$

Use the distributive property to multiply 3x-3 by C.

$$3xC-3C=y\left(-2x+4\right)$$

Use the distributive property to multiply y by -2x+4.

$$3xC-3C=-2yx+4y$$

Add 2yx to both sides.

$$3xC-3C+2yx=4y$$

Add 3C to both sides.

$$3xC+2yx=4y+3C$$

Combine all terms containing x.

$$\left(3C+2y\right)x=4y+3C$$

The equation is in standard form.

$$\left(2y+3C\right)x=4y+3C$$

Divide both sides by 2y+3C.

$$\frac{\left(2y+3C\right)x}{2y+3C}=\frac{4y+3C}{2y+3C}$$

Dividing by 2y+3C undoes the multiplication by 2y+3C.

$$x=\frac{4y+3C}{2y+3C}$$

Variable x cannot be equal to 1.

$$x=\frac{4y+3C}{2y+3C}\text{, }x\neq 1$$