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

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

Solve for A

$$A=2\sqrt{3}+1\approx 4.464101615$$

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.

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

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

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

Subtract 1 from 2 to get 1.

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

Assign A

$$A≔2\sqrt{3}+1$$