Solve 2*frac{8/9+frac{2sqrt{3}}{9}}{frac{2sqrt{2}}{3}-1/3}

$$\frac{ 2 \times \frac{ 8 }{ 9 } + \frac{ 2 \sqrt{ 3 } }{ 9 } }{ \frac{ 2 \sqrt{ 2 } }{ 3 } - \frac{ 1 }{ 3 } }$$

Evaluate

$$\frac{2\left(\sqrt{3}+2\sqrt{6}+16\sqrt{2}+8\right)}{21}\approx 3.548423552$$

Short Solution Steps

Express 2\times \frac{8}{9} as a single fraction.

$$\frac{\frac{2\times 8}{9}+\frac{2\sqrt{3}}{9}}{\frac{2\sqrt{2}}{3}-\frac{1}{3}}$$

Multiply 2 and 8 to get 16.

$$\frac{\frac{16}{9}+\frac{2\sqrt{3}}{9}}{\frac{2\sqrt{2}}{3}-\frac{1}{3}}$$

Since \frac{16}{9} and \frac{2\sqrt{3}}{9} have the same denominator, add them by adding their numerators.

$$\frac{\frac{16+2\sqrt{3}}{9}}{\frac{2\sqrt{2}}{3}-\frac{1}{3}}$$

Since \frac{2\sqrt{2}}{3} and \frac{1}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{16+2\sqrt{3}}{9}}{\frac{2\sqrt{2}-1}{3}}$$

Divide \frac{16+2\sqrt{3}}{9} by \frac{2\sqrt{2}-1}{3} by multiplying \frac{16+2\sqrt{3}}{9} by the reciprocal of \frac{2\sqrt{2}-1}{3}.

$$\frac{\left(16+2\sqrt{3}\right)\times 3}{9\left(2\sqrt{2}-1\right)}$$

Cancel out 3 in both numerator and denominator.

$$\frac{2\sqrt{3}+16}{3\left(2\sqrt{2}-1\right)}$$

Use the distributive property to multiply 3 by 2\sqrt{2}-1.

$$\frac{2\sqrt{3}+16}{6\sqrt{2}-3}$$

Rationalize the denominator of \frac{2\sqrt{3}+16}{6\sqrt{2}-3} by multiplying numerator and denominator by 6\sqrt{2}+3.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{\left(6\sqrt{2}-3\right)\left(6\sqrt{2}+3\right)}$$

Consider \left(6\sqrt{2}-3\right)\left(6\sqrt{2}+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}.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{\left(6\sqrt{2}\right)^{2}-3^{2}}$$

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

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{6^{2}\left(\sqrt{2}\right)^{2}-3^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{36\left(\sqrt{2}\right)^{2}-3^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{36\times 2-3^{2}}$$

Multiply 36 and 2 to get 72.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{72-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{72-9}$$

Subtract 9 from 72 to get 63.

$$\frac{\left(2\sqrt{3}+16\right)\left(6\sqrt{2}+3\right)}{63}$$

Apply the distributive property by multiplying each term of 2\sqrt{3}+16 by each term of 6\sqrt{2}+3.

$$\frac{12\sqrt{3}\sqrt{2}+6\sqrt{3}+96\sqrt{2}+48}{63}$$

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

$$\frac{12\sqrt{6}+6\sqrt{3}+96\sqrt{2}+48}{63}$$

Factor

$$\frac{2 {(\sqrt{3} + 2 \sqrt{6} + 16 \sqrt{2} + 8)}}{21} = 3.5484235515337863$$