Solve a=frac{8sqrt{6}*2sqrt{3}}{5sqrt{6}-3sqrt{2}}

$$a = \frac { 8 \sqrt { 6 } \cdot 2 \sqrt { 3 } } { 5 \sqrt { 6 } - 3 \sqrt { 2 } }$$

Solve for a

$$a = \frac{40 \sqrt{3} + 24}{11} \approx 8.480184755$$

Steps for Solving Linear Equation

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}.

$$a=\frac{2\times 8\sqrt{3}\sqrt{2}\sqrt{3}}{5\sqrt{6}-3\sqrt{2}}$$

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

$$a=\frac{2\times 8\times 3\sqrt{2}}{5\sqrt{6}-3\sqrt{2}}$$

Multiply 2 and 8 to get 16.

$$a=\frac{16\times 3\sqrt{2}}{5\sqrt{6}-3\sqrt{2}}$$

Multiply 16 and 3 to get 48.

$$a=\frac{48\sqrt{2}}{5\sqrt{6}-3\sqrt{2}}$$

Rationalize the denominator of \frac{48\sqrt{2}}{5\sqrt{6}-3\sqrt{2}} by multiplying numerator and denominator by 5\sqrt{6}+3\sqrt{2}.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{\left(5\sqrt{6}-3\sqrt{2}\right)\left(5\sqrt{6}+3\sqrt{2}\right)}$$

Consider \left(5\sqrt{6}-3\sqrt{2}\right)\left(5\sqrt{6}+3\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{\left(5\sqrt{6}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}$$

Expand \left(5\sqrt{6}\right)^{2}.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{5^{2}\left(\sqrt{6}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{25\left(\sqrt{6}\right)^{2}-\left(-3\sqrt{2}\right)^{2}}$$

The square of \sqrt{6} is 6.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{25\times 6-\left(-3\sqrt{2}\right)^{2}}$$

Multiply 25 and 6 to get 150.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{150-\left(-3\sqrt{2}\right)^{2}}$$

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

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{150-\left(-3\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -3 to the power of 2 and get 9.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{150-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{150-9\times 2}$$

Multiply 9 and 2 to get 18.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{150-18}$$

Subtract 18 from 150 to get 132.

$$a=\frac{48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)}{132}$$

Divide 48\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right) by 132 to get \frac{4}{11}\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right).

$$a=\frac{4}{11}\sqrt{2}\left(5\sqrt{6}+3\sqrt{2}\right)$$

Use the distributive property to multiply \frac{4}{11}\sqrt{2} by 5\sqrt{6}+3\sqrt{2}.

$$a=\frac{4}{11}\sqrt{2}\times 5\sqrt{6}+\frac{4}{11}\sqrt{2}\times 3\sqrt{2}$$

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

$$a=\frac{4}{11}\sqrt{2}\times 5\sqrt{6}+\frac{4}{11}\times 2\times 3$$

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}.

$$a=\frac{4}{11}\sqrt{2}\times 5\sqrt{2}\sqrt{3}+\frac{4}{11}\times 2\times 3$$

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

$$a=\frac{4}{11}\times 2\times 5\sqrt{3}+\frac{4}{11}\times 2\times 3$$

Express \frac{4}{11}\times 2 as a single fraction.

$$a=\frac{4\times 2}{11}\times 5\sqrt{3}+\frac{4}{11}\times 2\times 3$$

Multiply 4 and 2 to get 8.

$$a=\frac{8}{11}\times 5\sqrt{3}+\frac{4}{11}\times 2\times 3$$

Express \frac{8}{11}\times 5 as a single fraction.

$$a=\frac{8\times 5}{11}\sqrt{3}+\frac{4}{11}\times 2\times 3$$

Multiply 8 and 5 to get 40.

$$a=\frac{40}{11}\sqrt{3}+\frac{4}{11}\times 2\times 3$$

Express \frac{4}{11}\times 2 as a single fraction.

$$a=\frac{40}{11}\sqrt{3}+\frac{4\times 2}{11}\times 3$$

Multiply 4 and 2 to get 8.

$$a=\frac{40}{11}\sqrt{3}+\frac{8}{11}\times 3$$

Express \frac{8}{11}\times 3 as a single fraction.

$$a=\frac{40}{11}\sqrt{3}+\frac{8\times 3}{11}$$

Multiply 8 and 3 to get 24.

$$a=\frac{40}{11}\sqrt{3}+\frac{24}{11}$$

Assign a

$$a≔\frac{40\sqrt{3}+24}{11}$$