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

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

Solve for b

$$b=-\frac{\sqrt{3}\left(a-4\sqrt{3}-7\right)}{3}$$

Steps for Solving Linear Equation

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

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

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

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{3} is 3.

$$4+4\sqrt{3}+3=a+b\sqrt{3}$$

Add 4 and 3 to get 7.

$$7+4\sqrt{3}=a+b\sqrt{3}$$

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

$$a+b\sqrt{3}=7+4\sqrt{3}$$

Subtract a from both sides.

$$b\sqrt{3}=7+4\sqrt{3}-a$$

The equation is in standard form.

$$\sqrt{3}b=-a+4\sqrt{3}+7$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+4\sqrt{3}+7}{\sqrt{3}}$$

Dividing by \sqrt{3} undoes the multiplication by \sqrt{3}.

$$b=\frac{-a+4\sqrt{3}+7}{\sqrt{3}}$$

Divide 4\sqrt{3}-a+7 by \sqrt{3}.

$$b=\frac{\sqrt{3}\left(-a+4\sqrt{3}+7\right)}{3}$$

Solve for a

$$a=-\sqrt{3}b+4\sqrt{3}+7$$