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

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

Solve for b

$$b=-\frac{\sqrt{2}a}{2}+\frac{11\sqrt{2}}{14}+\frac{6}{7}$$

Steps for Solving Linear Equation

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

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

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

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

Square 3. Square \sqrt{2}.

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

Subtract 2 from 9 to get 7.

$$\frac{\left(3+\sqrt{2}\right)\left(3+\sqrt{2}\right)}{7}=a+b\sqrt{2}$$

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

$$\frac{\left(3+\sqrt{2}\right)^{2}}{7}=a+b\sqrt{2}$$

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

$$\frac{9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}}{7}=a+b\sqrt{2}$$

The square of \sqrt{2} is 2.

$$\frac{9+6\sqrt{2}+2}{7}=a+b\sqrt{2}$$

Add 9 and 2 to get 11.

$$\frac{11+6\sqrt{2}}{7}=a+b\sqrt{2}$$

Divide each term of 11+6\sqrt{2} by 7 to get \frac{11}{7}+\frac{6}{7}\sqrt{2}.

$$\frac{11}{7}+\frac{6}{7}\sqrt{2}=a+b\sqrt{2}$$

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

$$a+b\sqrt{2}=\frac{11}{7}+\frac{6}{7}\sqrt{2}$$

Subtract a from both sides.

$$b\sqrt{2}=\frac{11}{7}+\frac{6}{7}\sqrt{2}-a$$

The equation is in standard form.

$$\sqrt{2}b=-a+\frac{6\sqrt{2}}{7}+\frac{11}{7}$$

Divide both sides by \sqrt{2}.

$$\frac{\sqrt{2}b}{\sqrt{2}}=\frac{-a+\frac{6\sqrt{2}}{7}+\frac{11}{7}}{\sqrt{2}}$$

Dividing by \sqrt{2} undoes the multiplication by \sqrt{2}.

$$b=\frac{-a+\frac{6\sqrt{2}}{7}+\frac{11}{7}}{\sqrt{2}}$$

Divide \frac{6\sqrt{2}}{7}-a+\frac{11}{7} by \sqrt{2}.

$$b=-\frac{\sqrt{2}a}{2}+\frac{11\sqrt{2}}{14}+\frac{6}{7}$$

Solve for a

$$a=-\sqrt{2}b+\frac{6\sqrt{2}}{7}+\frac{11}{7}$$