Solve frac{-1+sqrt{3}}{1+sqrt{3}}-sqrt{3}+b=1

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

Solve for b

$$b=2\sqrt{3}-1\approx 2.464101615$$

Steps for Solving Linear Equation

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

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

Consider \left(1+\sqrt{3}\right)\left(1-\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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

Apply the distributive property by multiplying each term of -1+\sqrt{3} by each term of 1-\sqrt{3}.

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

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from -1 to get -4.

$$\frac{-4+2\sqrt{3}}{-2}-\sqrt{3}+b=1$$

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

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

Combine -\sqrt{3} and -\sqrt{3} to get -2\sqrt{3}.

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

Subtract 2 from both sides.

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

Subtract 2 from 1 to get -1.

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

Add 2\sqrt{3} to both sides.

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