Solve frac{sqrt{3}}{2sqrt{3}-3}=x+ysqrt{3}

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

Solve for y

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

Steps for Solving Linear Equation

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{3}-3} by multiplying numerator and denominator by 2\sqrt{3}+3.

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

Consider \left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(2\sqrt{3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{\sqrt{3}\left(2\sqrt{3}+3\right)}{12-3^{2}}=x+y\sqrt{3}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\sqrt{3}\left(2\sqrt{3}+3\right)}{12-9}=x+y\sqrt{3}$$

Subtract 9 from 12 to get 3.

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

Use the distributive property to multiply \sqrt{3} by 2\sqrt{3}+3.

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

Divide each term of 6+3\sqrt{3} by 3 to get 2+\sqrt{3}.

$$2+\sqrt{3}=x+y\sqrt{3}$$

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

$$x+y\sqrt{3}=2+\sqrt{3}$$

Subtract x from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

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

Divide \sqrt{3}+2-x by \sqrt{3}.

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

Solve for x

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