Solve 3/sqrt{2}+x/2sqrt{3}=1

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

Solve for x

$$x=2\sqrt{3}-3\sqrt{6}\approx -3.884367613$$

Steps for Solving Linear Equation

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

$$\frac{3\sqrt{2}}{2}+\frac{x\sqrt{3}}{6}=1$$

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

$$\frac{3\times 3\sqrt{2}}{6}+\frac{x\sqrt{3}}{6}=1$$

Since \frac{3\times 3\sqrt{2}}{6} and \frac{x\sqrt{3}}{6} have the same denominator, add them by adding their numerators.

$$\frac{3\times 3\sqrt{2}+x\sqrt{3}}{6}=1$$

Do the multiplications in 3\times 3\sqrt{2}+x\sqrt{3}.

$$\frac{9\sqrt{2}+x\sqrt{3}}{6}=1$$

Multiply both sides by 6.

$$9\sqrt{2}+x\sqrt{3}=6$$

Subtract 9\sqrt{2} from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

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

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

Divide 6-9\sqrt{2} by \sqrt{3}.

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