Solve frac{sqrt{6}}{sqrt{2}tsqrt{3}}=frac{sqrt{6}(sqrt{2}-sqrt{3})}{(sqrt{2}+sqrt{3})(sqrt{2}-sqrt{3})}=

$$\frac { \sqrt { 6 } } { \sqrt { 2 } t \sqrt { 3 } } = \frac { \sqrt { 6 } ( \sqrt { 2 } - \sqrt { 3 } ) } { ( \sqrt { 2 } + \sqrt { 3 } ) ( \sqrt { 2 } - \sqrt { 3 } ) } =$$

Solve for t

$$t = \frac{2 \sqrt{3} + 3 \sqrt{2}}{6} \approx 1.28445705$$

Steps for Solving Linear Equation

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

Rationalize the denominator of \frac{\sqrt{6}}{\sqrt{6}t} by multiplying numerator and denominator by \sqrt{6}.

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

The square of \sqrt{6} is 6.

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

Multiply \sqrt{6} and \sqrt{6} to get 6.

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

Consider \left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\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}.

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

Square \sqrt{2}. Square \sqrt{3}.

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

Subtract 3 from 2 to get -1.

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

Anything divided by -1 gives its opposite.

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

Use the distributive property to multiply \sqrt{6} by \sqrt{2}-\sqrt{3}.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

To find the opposite of 2\sqrt{3}-3\sqrt{2}, find the opposite of each term.

$$\frac{6}{6t}=-2\sqrt{3}+3\sqrt{2}$$

Variable t cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 6t.

$$6=-2\sqrt{3}\times 6t+3\sqrt{2}\times 6t$$

Reorder the terms.

$$6=3\times 6\sqrt{2}t-2\times 6\sqrt{3}t$$

Do the multiplications.

$$6=18\sqrt{2}t-12\sqrt{3}t$$

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

$$18\sqrt{2}t-12\sqrt{3}t=6$$

Combine all terms containing t.

$$\left(18\sqrt{2}-12\sqrt{3}\right)t=6$$

Divide both sides by 18\sqrt{2}-12\sqrt{3}.

$$\frac{\left(18\sqrt{2}-12\sqrt{3}\right)t}{18\sqrt{2}-12\sqrt{3}}=\frac{6}{18\sqrt{2}-12\sqrt{3}}$$

Dividing by 18\sqrt{2}-12\sqrt{3} undoes the multiplication by 18\sqrt{2}-12\sqrt{3}.

$$t=\frac{6}{18\sqrt{2}-12\sqrt{3}}$$

Divide 6 by 18\sqrt{2}-12\sqrt{3}.

$$t=\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{3}$$