Solve -frac{2sqrt{3}}{3}+frac{sqrt{2}}{2}

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

Evaluate

$$\frac{\sqrt{3}\left(\sqrt{6}-4\right)}{6}\approx -0.447593757$$

Short Solution Steps

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

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

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

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

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

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

Factor

$$\frac{\sqrt{3} {(\sqrt{2} \sqrt{3} - 4)}}{6} = -0.44759375719270394$$