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

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

Solve for y (complex solution)

$$\left\{\begin{matrix}y=-\frac{3x-1}{2x-1}\text{, }&x\neq \frac{1}{2}\\y\in \mathrm{C}\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

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

$$2yx^{2}+3x^{2}+yx+2x-y-1=0$$

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

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

Subtract 2x from both sides.

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

Add 1 to both sides.

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

Combine all terms containing y.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for y

$$\left\{\begin{matrix}y=-\frac{3x-1}{2x-1}\text{, }&x\neq \frac{1}{2}\\y\in \mathrm{R}\text{, }&x=-1\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

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

$$2yx^{2}+3x^{2}+yx+2x-y-1=0$$

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

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

Subtract 2x from both sides.

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

Add 1 to both sides.

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

Combine all terms containing y.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for x

$$\left\{\begin{matrix}\\x=-1\text{, }&\text{unconditionally}\\x=\frac{y+1}{2y+3}\text{, }&y\neq -\frac{3}{2}\end{matrix}\right.$$