Solve y=x^4-2x^3+3x^2-3x-7-xy+8-xy

$$y = x ^ { 4 } - 2 x ^ { 3 } + 3 x ^ { 2 } - 3 x - 7 - x y + 8 - x y$$

Solve for y

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

$$x\neq -\frac{1}{2}$$

Steps for Solving Linear Equation

Add -7 and 8 to get 1.

$$y=x^{4}-2x^{3}+3x^{2}-3x+1-xy-xy$$

Combine -xy and -xy to get -2xy.

$$y=x^{4}-2x^{3}+3x^{2}-3x+1-2xy$$

Add 2xy to both sides.

$$y+2xy=x^{4}-2x^{3}+3x^{2}-3x+1$$

Combine all terms containing y.

$$\left(1+2x\right)y=x^{4}-2x^{3}+3x^{2}-3x+1$$

The equation is in standard form.

$$\left(2x+1\right)y=x^{4}-2x^{3}+3x^{2}-3x+1$$

Divide both sides by 1+2x.

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

Dividing by 1+2x undoes the multiplication by 1+2x.

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

Divide \left(-1+x\right)\left(x^{3}-\left(x-1\right)^{2}\right) by 1+2x.

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