Solve x<frac{6-2sqrt{3}}{2sqrt{3}}

$$x < \frac { 6 - 2 \sqrt { 3 } } { 2 \sqrt { 3 } }$$

Solve for x

$$x<\sqrt{3}-1$$

Solution Steps

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

$$x<\frac{\left(6-2\sqrt{3}\right)\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$x<\frac{\left(6-2\sqrt{3}\right)\sqrt{3}}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$x<\frac{\left(6-2\sqrt{3}\right)\sqrt{3}}{6}$$

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

$$x<\frac{6\sqrt{3}-2\left(\sqrt{3}\right)^{2}}{6}$$

The square of \sqrt{3} is 3.

$$x<\frac{6\sqrt{3}-2\times 3}{6}$$

Multiply -2 and 3 to get -6.

$$x<\frac{6\sqrt{3}-6}{6}$$

Divide each term of 6\sqrt{3}-6 by 6 to get \sqrt{3}-1.

$$x<\sqrt{3}-1$$