Solve 5x^2-3xy+x+y=0

$$5 x ^ { 2 } - 3 x y + x + y = 0$$

Solve for y

$$y=-\frac{x\left(5x+1\right)}{1-3x}$$

$$x\neq \frac{1}{3}$$

Steps for Solving Linear Equation

Subtract 5x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-3xy+x+y=-5x^{2}$$

Subtract x from both sides.

$$-3xy+y=-5x^{2}-x$$

Combine all terms containing y.

$$\left(-3x+1\right)y=-5x^{2}-x$$

The equation is in standard form.

$$\left(1-3x\right)y=-5x^{2}-x$$

Divide both sides by -3x+1.

$$\frac{\left(1-3x\right)y}{1-3x}=-\frac{x\left(5x+1\right)}{1-3x}$$

Dividing by -3x+1 undoes the multiplication by -3x+1.

$$y=-\frac{x\left(5x+1\right)}{1-3x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{9y^{2}-26y+1}+3y-1}{10}$$

$$x=\frac{-\sqrt{9y^{2}-26y+1}+3y-1}{10}$$

Solve for x

$$x=\frac{\sqrt{9y^{2}-26y+1}+3y-1}{10}$$

$$x=\frac{-\sqrt{9y^{2}-26y+1}+3y-1}{10}\text{, }y\geq \frac{4\sqrt{10}+13}{9}\text{ or }y\leq \frac{13-4\sqrt{10}}{9}$$