Solve y=3x-1/8x-3

$$y = \frac { 3 x - 1 } { 8 x - 3 }$$

Solve for x

$$x=-\frac{1-3y}{8y-3}$$

$$y\neq \frac{3}{8}$$

Steps for Solving Linear Equation

Variable x cannot be equal to \frac{3}{8} since division by zero is not defined. Multiply both sides of the equation by 8x-3.

$$y\left(8x-3\right)=3x-1$$

Use the distributive property to multiply y by 8x-3.

$$8yx-3y=3x-1$$

Subtract 3x from both sides.

$$8yx-3y-3x=-1$$

Add 3y to both sides.

$$8yx-3x=-1+3y$$

Combine all terms containing x.

$$\left(8y-3\right)x=-1+3y$$

The equation is in standard form.

$$\left(8y-3\right)x=3y-1$$

Divide both sides by 8y-3.

$$\frac{\left(8y-3\right)x}{8y-3}=\frac{3y-1}{8y-3}$$

Dividing by 8y-3 undoes the multiplication by 8y-3.

$$x=\frac{3y-1}{8y-3}$$

Variable x cannot be equal to \frac{3}{8}.

$$x=\frac{3y-1}{8y-3}\text{, }x\neq \frac{3}{8}$$

Solve for y

$$y=-\frac{1-3x}{8x-3}$$

$$x\neq \frac{3}{8}$$