Solve (y+3)x^2+(2y+1)x+y=0

$$(y+3) { x }^{ 2 } +(2y+1)x+y=0$$

Solve for y

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

$$x\neq -1$$

Steps for Solving Linear Equation

Use the distributive property to multiply y+3 by x^{2}.

$$yx^{2}+3x^{2}+\left(2y+1\right)x+y=0$$

Use the distributive property to multiply 2y+1 by x.

$$yx^{2}+3x^{2}+2yx+x+y=0$$

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$yx^{2}+2yx+x+y=-3x^{2}$$

Subtract x from both sides.

$$yx^{2}+2yx+y=-3x^{2}-x$$

Combine all terms containing y.

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

Divide both sides by x^{2}+2x+1.

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

Dividing by x^{2}+2x+1 undoes the multiplication by x^{2}+2x+1.

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

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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{-2y+\sqrt{1-8y}-1}{2\left(y+3\right)}\text{; }x=-\frac{2y+\sqrt{1-8y}+1}{2\left(y+3\right)}\text{, }&y\neq -3\\x=-\frac{3}{5}=-0.6\text{, }&y=-3\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-2y+\sqrt{1-8y}-1}{2\left(y+3\right)}\text{; }x=-\frac{2y+\sqrt{1-8y}+1}{2\left(y+3\right)}\text{, }&y\neq -3\text{ and }y\leq \frac{1}{8}\\x=-\frac{3}{5}=-0.6\text{, }&y=-3\end{matrix}\right.$$