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

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

Solve for b

$$b=-\frac{\sqrt{6}\left(a-2\sqrt{6}-5\right)}{6}$$

Steps for Solving Linear Equation

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

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

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}.

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

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

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

Subtract 2 from 3 to get 1.

$$\frac{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{1}=a+b\sqrt{6}$$

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

$$5+2\sqrt{6}=a+b\sqrt{6}$$

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

$$a+b\sqrt{6}=5+2\sqrt{6}$$

Subtract a from both sides.

$$b\sqrt{6}=5+2\sqrt{6}-a$$

The equation is in standard form.

$$\sqrt{6}b=-a+2\sqrt{6}+5$$

Divide both sides by \sqrt{6}.

$$\frac{\sqrt{6}b}{\sqrt{6}}=\frac{-a+2\sqrt{6}+5}{\sqrt{6}}$$

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

$$b=\frac{-a+2\sqrt{6}+5}{\sqrt{6}}$$

Divide -a+2\sqrt{6}+5 by \sqrt{6}.

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

Solve for a

$$a=-\sqrt{6}b+2\sqrt{6}+5$$