Solve frac{sqrt{3}+sqrt{2}}{3sqrt{3}-2sqrt{2}}=a+bsqrt{6}

$$\frac { \sqrt { 3 } + \sqrt { 2 } } { 3 \sqrt { 3 } - 2 \sqrt { 2 } } = a + b \sqrt { 6 }$$

Solve for b

$$b=-\frac{\sqrt{6}a}{6}+\frac{13\sqrt{6}}{114}+\frac{5}{19}$$

Steps for Solving Linear Equation

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

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

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

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

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

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{9\left(\sqrt{3}\right)^{2}-\left(-2\sqrt{2}\right)^{2}}=a+b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{9\times 3-\left(-2\sqrt{2}\right)^{2}}=a+b\sqrt{6}$$

Multiply 9 and 3 to get 27.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{27-\left(-2\sqrt{2}\right)^{2}}=a+b\sqrt{6}$$

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

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{27-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}=a+b\sqrt{6}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{27-4\left(\sqrt{2}\right)^{2}}=a+b\sqrt{6}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{27-4\times 2}=a+b\sqrt{6}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{27-8}=a+b\sqrt{6}$$

Subtract 8 from 27 to get 19.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(3\sqrt{3}+2\sqrt{2}\right)}{19}=a+b\sqrt{6}$$

Use the distributive property to multiply \sqrt{3}+\sqrt{2} by 3\sqrt{3}+2\sqrt{2} and combine like terms.

$$\frac{3\left(\sqrt{3}\right)^{2}+5\sqrt{3}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{19}=a+b\sqrt{6}$$

The square of \sqrt{3} is 3.

$$\frac{3\times 3+5\sqrt{3}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{19}=a+b\sqrt{6}$$

Multiply 3 and 3 to get 9.

$$\frac{9+5\sqrt{3}\sqrt{2}+2\left(\sqrt{2}\right)^{2}}{19}=a+b\sqrt{6}$$

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

$$\frac{9+5\sqrt{6}+2\left(\sqrt{2}\right)^{2}}{19}=a+b\sqrt{6}$$

The square of \sqrt{2} is 2.

$$\frac{9+5\sqrt{6}+2\times 2}{19}=a+b\sqrt{6}$$

Multiply 2 and 2 to get 4.

$$\frac{9+5\sqrt{6}+4}{19}=a+b\sqrt{6}$$

Add 9 and 4 to get 13.

$$\frac{13+5\sqrt{6}}{19}=a+b\sqrt{6}$$

Divide each term of 13+5\sqrt{6} by 19 to get \frac{13}{19}+\frac{5}{19}\sqrt{6}.

$$\frac{13}{19}+\frac{5}{19}\sqrt{6}=a+b\sqrt{6}$$

Swap sides so that all variable terms are on the left hand side.

$$a+b\sqrt{6}=\frac{13}{19}+\frac{5}{19}\sqrt{6}$$

Subtract a from both sides.

$$b\sqrt{6}=\frac{13}{19}+\frac{5}{19}\sqrt{6}-a$$

The equation is in standard form.

$$\sqrt{6}b=-a+\frac{5\sqrt{6}}{19}+\frac{13}{19}$$

Divide both sides by \sqrt{6}.

$$\frac{\sqrt{6}b}{\sqrt{6}}=\frac{-a+\frac{5\sqrt{6}}{19}+\frac{13}{19}}{\sqrt{6}}$$

Dividing by \sqrt{6} undoes the multiplication by \sqrt{6}.

$$b=\frac{-a+\frac{5\sqrt{6}}{19}+\frac{13}{19}}{\sqrt{6}}$$

Divide \frac{13}{19}+\frac{5\sqrt{6}}{19}-a by \sqrt{6}.

$$b=-\frac{\sqrt{6}a}{6}+\frac{13\sqrt{6}}{114}+\frac{5}{19}$$

Solve for a

$$a=-\sqrt{6}b+\frac{5\sqrt{6}}{19}+\frac{13}{19}$$