Solve frac{sqrt{3}+2}{sqrt{3}-2}=a+5sqrt{3}

$$\frac { \sqrt { 3 } + 2 } { \sqrt { 3 } - 2 } = a + 5 \sqrt { 3 }$$

Solve for a

$$a=-9\sqrt{3}-7\approx -22.588457268$$

Steps for Solving Linear Equation

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

$$\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}=a+5\sqrt{3}$$

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

$$\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}\right)^{2}-2^{2}}=a+5\sqrt{3}$$

Square \sqrt{3}. Square 2.

$$\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{3-4}=a+5\sqrt{3}$$

Subtract 4 from 3 to get -1.

$$\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{-1}=a+5\sqrt{3}$$

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

$$\frac{\left(\sqrt{3}+2\right)^{2}}{-1}=a+5\sqrt{3}$$

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

$$\frac{\left(\sqrt{3}\right)^{2}+4\sqrt{3}+4}{-1}=a+5\sqrt{3}$$

The square of \sqrt{3} is 3.

$$\frac{3+4\sqrt{3}+4}{-1}=a+5\sqrt{3}$$

Add 3 and 4 to get 7.

$$\frac{7+4\sqrt{3}}{-1}=a+5\sqrt{3}$$

Anything divided by -1 gives its opposite. To find the opposite of 7+4\sqrt{3}, find the opposite of each term.

$$-7-4\sqrt{3}=a+5\sqrt{3}$$

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

$$a+5\sqrt{3}=-7-4\sqrt{3}$$

Subtract 5\sqrt{3} from both sides.

$$a=-7-4\sqrt{3}-5\sqrt{3}$$

Combine -4\sqrt{3} and -5\sqrt{3} to get -9\sqrt{3}.

$$a=-7-9\sqrt{3}$$