Solve (5+2sqrt{3})/(7+4sqrt{3})=a+bsqrt{3}

$$( 5 + 2 \sqrt { 3 } ) / ( 7 + 4 \sqrt { 3 } ) = a + b \sqrt { 3 }$$

Solve for b

$$b=-\frac{\sqrt{3}\left(a+6\sqrt{3}-11\right)}{3}$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{5+2\sqrt{3}}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=a+b\sqrt{3}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\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}.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

The square of \sqrt{3} is 3.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\times 3}=a+b\sqrt{3}$$

Multiply 16 and 3 to get 48.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-48}=a+b\sqrt{3}$$

Subtract 48 from 49 to get 1.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{1}=a+b\sqrt{3}$$

Anything divided by one gives itself.

$$\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)=a+b\sqrt{3}$$

Use the distributive property to multiply 5+2\sqrt{3} by 7-4\sqrt{3} and combine like terms.

$$35-6\sqrt{3}-8\left(\sqrt{3}\right)^{2}=a+b\sqrt{3}$$

The square of \sqrt{3} is 3.

$$35-6\sqrt{3}-8\times 3=a+b\sqrt{3}$$

Multiply -8 and 3 to get -24.

$$35-6\sqrt{3}-24=a+b\sqrt{3}$$

Subtract 24 from 35 to get 11.

$$11-6\sqrt{3}=a+b\sqrt{3}$$

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

$$a+b\sqrt{3}=11-6\sqrt{3}$$

Subtract a from both sides.

$$b\sqrt{3}=11-6\sqrt{3}-a$$

The equation is in standard form.

$$\sqrt{3}b=-a+11-6\sqrt{3}$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+11-6\sqrt{3}}{\sqrt{3}}$$

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

$$b=\frac{-a+11-6\sqrt{3}}{\sqrt{3}}$$

Divide -6\sqrt{3}-a+11 by \sqrt{3}.

$$b=\frac{\sqrt{3}\left(-a+11-6\sqrt{3}\right)}{3}$$

Solve for a

$$a=-\sqrt{3}b+11-6\sqrt{3}$$