Solve frac{sqrt{3}-1}{sqrt{3}+1}=a+bsqrt{3}

$$\frac { \sqrt { 3 } - 1 } { \sqrt { 3 } + 1 } = a + b \sqrt { 3 }$$

Solve for b

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

Steps for Solving Linear Equation

Rationalize the denominator of \frac{\sqrt{3}-1}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Multiply \sqrt{3}-1 and \sqrt{3}-1 to get \left(\sqrt{3}-1\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-1\right)^{2}.

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

The square of \sqrt{3} is 3.

$$\frac{3-2\sqrt{3}+1}{2}=a+b\sqrt{3}$$

Add 3 and 1 to get 4.

$$\frac{4-2\sqrt{3}}{2}=a+b\sqrt{3}$$

Divide each term of 4-2\sqrt{3} by 2 to get 2-\sqrt{3}.

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

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

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

Subtract a from both sides.

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

The equation is in standard form.

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

Divide both sides by \sqrt{3}.

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

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

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

Divide -\sqrt{3}-a+2 by \sqrt{3}.

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

Solve for a

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