Solve f:frac{3+sqrt{7}}{3-sqrt{7}}=a+bsqrt{7}

$$f : \frac { 3 + \sqrt { 7 } } { 3 - \sqrt { 7 } } = a + b \sqrt { 7 }$$

Solve for b

$$b=-\frac{\sqrt{7}\left(3\sqrt{7}f+a-8f\right)}{7}$$

Steps for Solving Linear Equation

Divide f by \frac{3+\sqrt{7}}{3-\sqrt{7}} by multiplying f by the reciprocal of \frac{3+\sqrt{7}}{3-\sqrt{7}}.

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

Rationalize the denominator of \frac{f\left(3-\sqrt{7}\right)}{3+\sqrt{7}} by multiplying numerator and denominator by 3-\sqrt{7}.

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

Consider \left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 3. Square \sqrt{7}.

$$\frac{f\left(3-\sqrt{7}\right)\left(3-\sqrt{7}\right)}{9-7}=a+b\sqrt{7}$$

Subtract 7 from 9 to get 2.

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

Multiply 3-\sqrt{7} and 3-\sqrt{7} to get \left(3-\sqrt{7}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{7}\right)^{2}.

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

The square of \sqrt{7} is 7.

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

Add 9 and 7 to get 16.

$$\frac{f\left(16-6\sqrt{7}\right)}{2}=a+b\sqrt{7}$$

Use the distributive property to multiply f by 16-6\sqrt{7}.

$$\frac{16f-6f\sqrt{7}}{2}=a+b\sqrt{7}$$

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

$$a+b\sqrt{7}=\frac{16f-6f\sqrt{7}}{2}$$

Subtract a from both sides.

$$b\sqrt{7}=\frac{16f-6f\sqrt{7}}{2}-a$$

Multiply both sides of the equation by 2.

$$2b\sqrt{7}=16f-6f\sqrt{7}-2a$$

Reorder the terms.

$$2\sqrt{7}b=-6\sqrt{7}f-2a+16f$$

The equation is in standard form.

$$2\sqrt{7}b=-6\sqrt{7}f+16f-2a$$

Divide both sides by 2\sqrt{7}.

$$\frac{2\sqrt{7}b}{2\sqrt{7}}=\frac{-6\sqrt{7}f+16f-2a}{2\sqrt{7}}$$

Dividing by 2\sqrt{7} undoes the multiplication by 2\sqrt{7}.

$$b=\frac{-6\sqrt{7}f+16f-2a}{2\sqrt{7}}$$

Divide -6\sqrt{7}f-2a+16f by 2\sqrt{7}.

$$b=\frac{\sqrt{7}\left(-3\sqrt{7}f+8f-a\right)}{7}$$

Solve for a

$$a=-3\sqrt{7}f-\sqrt{7}b+8f$$