Solve x^2-6x+y=(x-z)(x-z)-1

$${ x }^{ 2 } -6x+y=(x-z)(x-z)-1$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{y-z^{2}+1}{2\left(z-3\right)}\text{, }&z\neq 3\\x\in \mathrm{R}\text{, }&y=8\text{ and }z=3\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply x-z and x-z to get \left(x-z\right)^{2}.

$$x^{2}-6x+y=\left(x-z\right)^{2}-1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-z\right)^{2}.

$$x^{2}-6x+y=x^{2}-2xz+z^{2}-1$$

Subtract x^{2} from both sides.

$$x^{2}-6x+y-x^{2}=-2xz+z^{2}-1$$

Combine x^{2} and -x^{2} to get 0.

$$-6x+y=-2xz+z^{2}-1$$

Add 2xz to both sides.

$$-6x+y+2xz=z^{2}-1$$

Subtract y from both sides.

$$-6x+2xz=z^{2}-1-y$$

Combine all terms containing x.

$$\left(-6+2z\right)x=z^{2}-1-y$$

The equation is in standard form.

$$\left(2z-6\right)x=z^{2}-y-1$$

Divide both sides by 2z-6.

$$\frac{\left(2z-6\right)x}{2z-6}=\frac{z^{2}-y-1}{2z-6}$$

Dividing by 2z-6 undoes the multiplication by 2z-6.

$$x=\frac{z^{2}-y-1}{2z-6}$$

Divide z^{2}-1-y by 2z-6.

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

Solve for y

$$y=-2xz+6x+z^{2}-1$$

Solution Steps

Multiply x-z and x-z to get \left(x-z\right)^{2}.

$$x^{2}-6x+y=\left(x-z\right)^{2}-1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-z\right)^{2}.

$$x^{2}-6x+y=x^{2}-2xz+z^{2}-1$$

Subtract x^{2} from both sides.

$$-6x+y=x^{2}-2xz+z^{2}-1-x^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$-6x+y=-2xz+z^{2}-1$$

Add 6x to both sides.

$$y=-2xz+z^{2}-1+6x$$