Solve 7/6x*6=9y^2+3y

$$\frac{ 7 }{ 6x \times 6 } =9 { y }^{ 2 } +3y$$

Solve for x

$$x=\frac{7}{108y\left(3y+1\right)}$$

$$y\neq -\frac{1}{3}\text{ and }y\neq 0$$

Steps for Solving Linear Equation

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 36x.

$$7=9y^{2}\times 36x+3y\times 36x$$

Multiply 9 and 36 to get 324.

$$7=324y^{2}x+3y\times 36x$$

Multiply 3 and 36 to get 108.

$$7=324y^{2}x+108yx$$

Swap sides so that all variable terms are on the left hand side.

$$324y^{2}x+108yx=7$$

Combine all terms containing x.

$$\left(324y^{2}+108y\right)x=7$$

Divide both sides by 324y^{2}+108y.

$$\frac{\left(324y^{2}+108y\right)x}{324y^{2}+108y}=\frac{7}{324y^{2}+108y}$$

Dividing by 324y^{2}+108y undoes the multiplication by 324y^{2}+108y.

$$x=\frac{7}{324y^{2}+108y}$$

Divide 7 by 324y^{2}+108y.

$$x=\frac{7}{108y\left(3y+1\right)}$$

Variable x cannot be equal to 0.

$$x=\frac{7}{108y\left(3y+1\right)}\text{, }x\neq 0$$

Solve for y (complex solution)

$$y=-\frac{\sqrt{x\left(9x+7\right)}+3x}{18x}$$

$$y=\frac{\sqrt{x\left(9x+7\right)}-3x}{18x}\text{, }x\neq 0$$

Solve for y

$$y=-\frac{\sqrt{x\left(9x+7\right)}+3x}{18x}$$

$$y=\frac{\sqrt{x\left(9x+7\right)}-3x}{18x}\text{, }x\leq -\frac{7}{9}\text{ or }x>0$$