Solve (x+3)(y-3)-(y^2-4x)=9

$$( x + 3 ) ( y - 3 ) - ( y ^ { 2 } - 4 x ) = 9$$

Solve for x

$$x=-\frac{-y^{2}+3y-18}{y+1}$$

$$y\neq -1$$

Steps for Solving Linear Equation

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

$$xy-3x+3y-9-\left(y^{2}-4x\right)=9$$

To find the opposite of y^{2}-4x, find the opposite of each term.

$$xy-3x+3y-9-y^{2}+4x=9$$

Combine -3x and 4x to get x.

$$xy+x+3y-9-y^{2}=9$$

Subtract 3y from both sides.

$$xy+x-9-y^{2}=9-3y$$

Add 9 to both sides.

$$xy+x-y^{2}=9-3y+9$$

Add 9 and 9 to get 18.

$$xy+x-y^{2}=18-3y$$

Add y^{2} to both sides.

$$xy+x=18-3y+y^{2}$$

Combine all terms containing x.

$$\left(y+1\right)x=18-3y+y^{2}$$

The equation is in standard form.

$$\left(y+1\right)x=y^{2}-3y+18$$

Divide both sides by y+1.

$$\frac{\left(y+1\right)x}{y+1}=\frac{y^{2}-3y+18}{y+1}$$

Dividing by y+1 undoes the multiplication by y+1.

$$x=\frac{y^{2}-3y+18}{y+1}$$

Solve for y (complex solution)

$$y=\frac{-\sqrt{x^{2}+10x-63}+x+3}{2}$$

$$y=\frac{\sqrt{x^{2}+10x-63}+x+3}{2}$$

Solve for y

$$y=\frac{-\sqrt{x^{2}+10x-63}+x+3}{2}$$

$$y=\frac{\sqrt{x^{2}+10x-63}+x+3}{2}\text{, }x\geq 2\sqrt{22}-5\text{ or }x\leq -2\sqrt{22}-5$$