Solve yx^2=-xy+6y+3

$$y { x }^{ 2 } =-xy+6y+3$$

Solve for y

$$y=\frac{3}{x^{2}+x-6}$$

$$x\neq -3\text{ and }x\neq 2$$

Steps for Solving Linear Equation

Subtract \left(-x\right)y from both sides.

$$yx^{2}-\left(-x\right)y=6y+3$$

Subtract 6y from both sides.

$$yx^{2}-\left(-x\right)y-6y=3$$

Multiply -1 and -1 to get 1.

$$yx^{2}+xy-6y=3$$

Combine all terms containing y.

$$\left(x^{2}+x-6\right)y=3$$

Divide both sides by x^{2}+x-6.

$$\frac{\left(x^{2}+x-6\right)y}{x^{2}+x-6}=\frac{3}{x^{2}+x-6}$$

Dividing by x^{2}+x-6 undoes the multiplication by x^{2}+x-6.

$$y=\frac{3}{x^{2}+x-6}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{25y^{2}+12y}}{2y}-\frac{1}{2}$$

$$x=-\frac{\sqrt{25y^{2}+12y}}{2y}-\frac{1}{2}\text{, }y\neq 0$$

Solve for x

$$x=\frac{\sqrt{25y^{2}+12y}}{2y}-\frac{1}{2}$$

$$x=-\frac{\sqrt{25y^{2}+12y}}{2y}-\frac{1}{2}\text{, }y>0\text{ or }y\leq -\frac{12}{25}$$