Solve y=2x-3/3x+3

$$y = \frac { 2 x - 3 } { 3 x + 3 }$$

Solve for x

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

$$y\neq \frac{2}{3}$$

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 3\left(x+1\right).

$$y\times 3\left(x+1\right)=2x-3$$

Use the distributive property to multiply y\times 3 by x+1.

$$3yx+y\times 3=2x-3$$

Subtract 2x from both sides.

$$3yx+y\times 3-2x=-3$$

Subtract y\times 3 from both sides.

$$3yx-2x=-3-y\times 3$$

Multiply -1 and 3 to get -3.

$$3yx-2x=-3-3y$$

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by 3y-2.

$$\frac{\left(3y-2\right)x}{3y-2}=\frac{-3y-3}{3y-2}$$

Dividing by 3y-2 undoes the multiplication by 3y-2.

$$x=\frac{-3y-3}{3y-2}$$

Divide -3-3y by 3y-2.

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

Variable x cannot be equal to -1.

$$x=-\frac{3\left(y+1\right)}{3y-2}\text{, }x\neq -1$$

Solve for y

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

$$x\neq -1$$