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

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

Solve for x

$$x=\frac{-3\sqrt{3}-9}{2}\approx -7.098076211$$

Steps for Solving Linear Equation

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

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

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}.

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

Square 3. Square \sqrt{3}.

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

Subtract 3 from 9 to get 6.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

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

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 in both numerator and denominator.

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

Use the distributive property to multiply 3x by 3-\sqrt{3}.

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

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

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

Reorder the terms.

$$-6\sqrt{3}x-3\sqrt{3}x+9x=3\times 9\sqrt{3}$$

Combine -6\sqrt{3}x and -3\sqrt{3}x to get -9\sqrt{3}x.

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

Multiply 3 and 9 to get 27.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

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

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

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

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