Solve frac{3sqrt{2}-2sqrt{3}}{3sqrt{2}+2sqrt{3}}+frac{2sqrt{3}}{sqrt{3}-sqrt{2}}=11

$$\frac { 3 \sqrt { 2 } - 2 \sqrt { 3 } } { 3 \sqrt { 2 } + 2 \sqrt { 3 } } + \frac { 2 \sqrt { 3 } } { \sqrt { 3 } - \sqrt { 2 } } = 11$$

Verify

Solution Steps

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

$$\frac{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}{\left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

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

$$\frac{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}-2\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Multiply 3\sqrt{2}-2\sqrt{3} and 3\sqrt{2}-2\sqrt{3} to get \left(3\sqrt{2}-2\sqrt{3}\right)^{2}.

$$\frac{\left(3\sqrt{2}-2\sqrt{3}\right)^{2}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-2\sqrt{3}\right)^{2}.

$$\frac{9\left(\sqrt{2}\right)^{2}-12\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

The square of \sqrt{2} is 2.

$$\frac{9\times 2-12\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Multiply 9 and 2 to get 18.

$$\frac{18-12\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

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

$$\frac{18-12\sqrt{6}+4\left(\sqrt{3}\right)^{2}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

The square of \sqrt{3} is 3.

$$\frac{18-12\sqrt{6}+4\times 3}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Multiply 4 and 3 to get 12.

$$\frac{18-12\sqrt{6}+12}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Add 18 and 12 to get 30.

$$\frac{30-12\sqrt{6}}{\left(3\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

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

$$\frac{30-12\sqrt{6}}{3^{2}\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Calculate 3 to the power of 2 and get 9.

$$\frac{30-12\sqrt{6}}{9\left(\sqrt{2}\right)^{2}-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

The square of \sqrt{2} is 2.

$$\frac{30-12\sqrt{6}}{9\times 2-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Multiply 9 and 2 to get 18.

$$\frac{30-12\sqrt{6}}{18-\left(2\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

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

$$\frac{30-12\sqrt{6}}{18-2^{2}\left(\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Calculate 2 to the power of 2 and get 4.

$$\frac{30-12\sqrt{6}}{18-4\left(\sqrt{3}\right)^{2}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

The square of \sqrt{3} is 3.

$$\frac{30-12\sqrt{6}}{18-4\times 3}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Multiply 4 and 3 to get 12.

$$\frac{30-12\sqrt{6}}{18-12}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Subtract 12 from 18 to get 6.

$$\frac{30-12\sqrt{6}}{6}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

Divide each term of 30-12\sqrt{6} by 6 to get 5-2\sqrt{6}.

$$5-2\sqrt{6}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11$$

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

$$5-2\sqrt{6}+\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=11$$

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

$$5-2\sqrt{6}+\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}=11$$

Square \sqrt{3}. Square \sqrt{2}.

$$5-2\sqrt{6}+\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{3-2}=11$$

Subtract 2 from 3 to get 1.

$$5-2\sqrt{6}+\frac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{1}=11$$

Anything divided by one gives itself.

$$5-2\sqrt{6}+2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)=11$$

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

$$5-2\sqrt{6}+2\left(\sqrt{3}\right)^{2}+2\sqrt{3}\sqrt{2}=11$$

The square of \sqrt{3} is 3.

$$5-2\sqrt{6}+2\times 3+2\sqrt{3}\sqrt{2}=11$$

Multiply 2 and 3 to get 6.

$$5-2\sqrt{6}+6+2\sqrt{3}\sqrt{2}=11$$

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

$$5-2\sqrt{6}+6+2\sqrt{6}=11$$

Add 5 and 6 to get 11.

$$11-2\sqrt{6}+2\sqrt{6}=11$$

Combine -2\sqrt{6} and 2\sqrt{6} to get 0.

$$11=11$$

Compare 11 and 11.

$$\text{true}$$