Solve 2sqrt{5}+4sqrt{3}-frac{2sqrt{18}-sqrt{27}}{sqrt{3}}

$$2 \sqrt { 5 } + 4 \sqrt { 3 } - \frac { 2 \sqrt { 18 } - \sqrt { 27 } } { \sqrt { 3 } }$$

Evaluate

$$2\sqrt{5}+4\sqrt{3}+3-2\sqrt{6}\approx 9.5013597$$

Short Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2\sqrt{5}+4\sqrt{3}-\frac{2\times 3\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

Multiply 2 and 3 to get 6.

$$2\sqrt{5}+4\sqrt{3}-\frac{6\sqrt{2}-\sqrt{27}}{\sqrt{3}}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$2\sqrt{5}+4\sqrt{3}-\frac{6\sqrt{2}-3\sqrt{3}}{\sqrt{3}}$$

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

$$2\sqrt{5}+4\sqrt{3}-\frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 2\sqrt{5}+4\sqrt{3} times \frac{3}{3}.

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

Since \frac{3\left(2\sqrt{5}+4\sqrt{3}\right)}{3} and \frac{\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(2\sqrt{5}+4\sqrt{3}\right)-\left(6\sqrt{2}-3\sqrt{3}\right)\sqrt{3}.

$$\frac{6\sqrt{5}+12\sqrt{3}-6\sqrt{6}+9}{3}$$

Divide each term of 6\sqrt{5}+12\sqrt{3}-6\sqrt{6}+9 by 3 to get 2\sqrt{5}+4\sqrt{3}-2\sqrt{6}+3.

$$2\sqrt{5}+4\sqrt{3}-2\sqrt{6}+3$$