Solve {3sqrt{7x}-y=2-frac{sqrt{3}}{2}

$$\{ 3 \sqrt { 7 x } - y = 2 - \frac { \sqrt { 3 } } { 2 }$$

Solve for x

$$x=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{252}$$

$$y\geq \frac{\sqrt{3}}{2}-2$$

Steps for Solving Radical Equations

Subtract -y from both sides of the equation.

$$3\sqrt{7x}-y-\left(-y\right)=-\frac{\sqrt{3}}{2}+2-\left(-y\right)$$

Subtracting -y from itself leaves 0.

$$3\sqrt{7x}=-\frac{\sqrt{3}}{2}+2-\left(-y\right)$$

Subtract -y from 2-\frac{1}{2}\sqrt{3}.

$$3\sqrt{7x}=y-\frac{\sqrt{3}}{2}+2$$

Divide both sides by 3.

$$\frac{3\sqrt{7x}}{3}=\frac{y-\frac{\sqrt{3}}{2}+2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$\sqrt{7x}=\frac{y-\frac{\sqrt{3}}{2}+2}{3}$$

Divide 2-\frac{\sqrt{3}}{2}+y by 3.

$$\sqrt{7x}=\frac{y}{3}-\frac{\sqrt{3}}{6}+\frac{2}{3}$$

Square both sides of the equation.

$$7x=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{36}$$

Divide both sides by 7.

$$\frac{7x}{7}=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{7\times 36}$$

Dividing by 7 undoes the multiplication by 7.

$$x=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{7\times 36}$$

Divide \frac{\left(4-\sqrt{3}+2y\right)^{2}}{36} by 7.

$$x=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{252}$$

Solve for x (complex solution)

$$x=\frac{\left(2y+4-\sqrt{3}\right)^{2}}{252}$$

$$arg(\frac{y}{21}-\frac{\sqrt{3}}{42}+\frac{2}{21})<\pi \text{ or }y=\frac{\sqrt{3}}{2}-2$$