Solve 3x/3+sqrt{3}-frac{2xsqrt{3}}{sqrt{3}-1}=27/2sqrt{3}

$$\frac { 3 x } { 3 + \sqrt { 3 } } - \frac { 2 x \sqrt { 3 } } { \sqrt { 3 } - 1 } = \frac { 27 } { 2 \sqrt { 3 } }$$

Solve for x

$$x=\frac{3\left(\sqrt{3}-3\right)}{2}\approx -1.901923789$$

Steps for Solving Linear Equation

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

$$\frac{3x\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}-\frac{2x\sqrt{3}}{\sqrt{3}-1}=\frac{27}{2\sqrt{3}}$$

Consider \left(3+\sqrt{3}\right)\left(3-\sqrt{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}.

$$\frac{3x\left(3-\sqrt{3}\right)}{3^{2}-\left(\sqrt{3}\right)^{2}}-\frac{2x\sqrt{3}}{\sqrt{3}-1}=\frac{27}{2\sqrt{3}}$$

Square 3. Square \sqrt{3}.

$$\frac{3x\left(3-\sqrt{3}\right)}{9-3}-\frac{2x\sqrt{3}}{\sqrt{3}-1}=\frac{27}{2\sqrt{3}}$$

Subtract 3 from 9 to get 6.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{2x\sqrt{3}}{\sqrt{3}-1}=\frac{27}{2\sqrt{3}}$$

Rationalize the denominator of \frac{2x\sqrt{3}}{\sqrt{3}-1} by multiplying numerator and denominator by \sqrt{3}+1.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{2x\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{27}{2\sqrt{3}}$$

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{2x\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}=\frac{27}{2\sqrt{3}}$$

Square \sqrt{3}. Square 1.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{2x\sqrt{3}\left(\sqrt{3}+1\right)}{3-1}=\frac{27}{2\sqrt{3}}$$

Subtract 1 from 3 to get 2.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{2x\sqrt{3}\left(\sqrt{3}+1\right)}{2}=\frac{27}{2\sqrt{3}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 6 and 2 is 6. Multiply \frac{2x\sqrt{3}\left(\sqrt{3}+1\right)}{2} times \frac{3}{3}.

$$\frac{3x\left(3-\sqrt{3}\right)}{6}-\frac{3\times 2x\sqrt{3}\left(\sqrt{3}+1\right)}{6}=\frac{27}{2\sqrt{3}}$$

Since \frac{3x\left(3-\sqrt{3}\right)}{6} and \frac{3\times 2x\sqrt{3}\left(\sqrt{3}+1\right)}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3x\left(3-\sqrt{3}\right)-3\times 2x\sqrt{3}\left(\sqrt{3}+1\right)}{6}=\frac{27}{2\sqrt{3}}$$

Do the multiplications in 3x\left(3-\sqrt{3}\right)-3\times 2x\sqrt{3}\left(\sqrt{3}+1\right).

$$\frac{9x-3x\sqrt{3}-18x-6x\sqrt{3}}{6}=\frac{27}{2\sqrt{3}}$$

Combine like terms in 9x-3x\sqrt{3}-18x-6x\sqrt{3}.

$$\frac{-9x-9x\sqrt{3}}{6}=\frac{27}{2\sqrt{3}}$$

Rationalize the denominator of \frac{27}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{-9x-9x\sqrt{3}}{6}=\frac{27\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{-9x-9x\sqrt{3}}{6}=\frac{27\sqrt{3}}{2\times 3}$$

Cancel out 3 in both numerator and denominator.

$$\frac{-9x-9x\sqrt{3}}{6}=\frac{9\sqrt{3}}{2}$$

Multiply both sides of the equation by 6, the least common multiple of 6,2.

$$-9x-9x\sqrt{3}=3\times 9\sqrt{3}$$

Reorder the terms.

$$-9\sqrt{3}x-9x=3\times 9\sqrt{3}$$

Multiply 3 and 9 to get 27.

$$-9\sqrt{3}x-9x=27\sqrt{3}$$

Combine all terms containing x.

$$\left(-9\sqrt{3}-9\right)x=27\sqrt{3}$$

Divide both sides by -9\sqrt{3}-9.

$$\frac{\left(-9\sqrt{3}-9\right)x}{-9\sqrt{3}-9}=\frac{27\sqrt{3}}{-9\sqrt{3}-9}$$

Dividing by -9\sqrt{3}-9 undoes the multiplication by -9\sqrt{3}-9.

$$x=\frac{27\sqrt{3}}{-9\sqrt{3}-9}$$

Divide 27\sqrt{3} by -9\sqrt{3}-9.

$$x=\frac{3\sqrt{3}-9}{2}$$