Solve y=8x-3/x+1

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

Solve for x

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

$$y\neq 8$$

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 x+1.

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

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

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

Subtract 8x from both sides.

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

Subtract y from both sides.

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

Combine all terms containing x.

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

The equation is in standard form.

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

Divide both sides by y-8.

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

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

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

Divide -3-y by y-8.

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

Variable x cannot be equal to -1.

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

Solve for y

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

$$x\neq -1$$