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

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

Evaluate

$$1$$

Short Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{2\sqrt{3}\left(2\sqrt{3}+3\right)}{12-3^{2}}+\frac{\sqrt{3}}{\sqrt{3}-2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{2\sqrt{3}\left(2\sqrt{3}+3\right)}{12-9}+\frac{\sqrt{3}}{\sqrt{3}-2}$$

Subtract 9 from 12 to get 3.

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

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

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

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

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{12+6\sqrt{3}}{3}-\sqrt{3}\left(\sqrt{3}+2\right)$$

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

$$\frac{12+6\sqrt{3}}{3}-\left(\left(\sqrt{3}\right)^{2}+2\sqrt{3}\right)$$

The square of \sqrt{3} is 3.

$$\frac{12+6\sqrt{3}}{3}-\left(3+2\sqrt{3}\right)$$

To find the opposite of 3+2\sqrt{3}, find the opposite of each term.

$$\frac{12+6\sqrt{3}}{3}-3-2\sqrt{3}$$

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

$$\frac{12+6\sqrt{3}}{3}+\frac{3\left(-3-2\sqrt{3}\right)}{3}$$

Since \frac{12+6\sqrt{3}}{3} and \frac{3\left(-3-2\sqrt{3}\right)}{3} have the same denominator, add them by adding their numerators.

$$\frac{12+6\sqrt{3}+3\left(-3-2\sqrt{3}\right)}{3}$$

Do the multiplications in 12+6\sqrt{3}+3\left(-3-2\sqrt{3}\right).

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

Do the calculations in 12+6\sqrt{3}-9-6\sqrt{3}.

$$\frac{3}{3}$$

Factor

$$1$$