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

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

Solve for b

$$b=2\left(\sqrt{3}-1\right)\approx 1.464101615$$

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}.

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

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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

Express \frac{\left(-1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{-2}\left(-\sqrt{3}\right) as a single fraction.

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

Use the distributive property to multiply -1 by -1+\sqrt{3}.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

$$\frac{\left(4-2\sqrt{3}\right)\sqrt{3}}{-2}+b=1$$

Use the distributive property to multiply 4-2\sqrt{3} by \sqrt{3}.

$$\frac{4\sqrt{3}-2\left(\sqrt{3}\right)^{2}}{-2}+b=1$$

The square of \sqrt{3} is 3.

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

Multiply -2 and 3 to get -6.

$$\frac{4\sqrt{3}-6}{-2}+b=1$$

Divide each term of 4\sqrt{3}-6 by -2 to get -2\sqrt{3}+3.

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

Add 2\sqrt{3} to both sides.

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

Subtract 3 from both sides.

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

Subtract 3 from 1 to get -2.

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