Solve frac{5+2sqrt{3}}{7+4sqrt{5}}=a+bsqrt{3}

$$\frac { 5 + 2 \sqrt { 3 } } { 7 + 4 \sqrt { 5 } } = a + b \sqrt { 3 }$$

Solve for b

$$b=-\frac{\sqrt{3}a}{3}+\frac{20\sqrt{15}}{93}+\frac{24\sqrt{5}}{93}-\frac{35\sqrt{3}}{93}-\frac{14}{31}$$

Steps for Solving Linear Equation

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

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

Consider \left(7+4\sqrt{5}\right)\left(7-4\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 7 to the power of 2 and get 49.

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

Expand \left(4\sqrt{5}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

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

The square of \sqrt{5} is 5.

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

Multiply 16 and 5 to get 80.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{5}\right)}{49-80}=a+b\sqrt{3}$$

Subtract 80 from 49 to get -31.

$$\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{5}\right)}{-31}=a+b\sqrt{3}$$

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

$$\frac{35-20\sqrt{5}+14\sqrt{3}-8\sqrt{3}\sqrt{5}}{-31}=a+b\sqrt{3}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{35-20\sqrt{5}+14\sqrt{3}-8\sqrt{15}}{-31}=a+b\sqrt{3}$$

Multiply both numerator and denominator by -1.

$$\frac{-35+20\sqrt{5}-14\sqrt{3}+8\sqrt{15}}{31}=a+b\sqrt{3}$$

Divide each term of -35+20\sqrt{5}-14\sqrt{3}+8\sqrt{15} by 31 to get -\frac{35}{31}+\frac{20}{31}\sqrt{5}-\frac{14}{31}\sqrt{3}+\frac{8}{31}\sqrt{15}.

$$-\frac{35}{31}+\frac{20}{31}\sqrt{5}-\frac{14}{31}\sqrt{3}+\frac{8}{31}\sqrt{15}=a+b\sqrt{3}$$

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

$$a+b\sqrt{3}=-\frac{35}{31}+\frac{20}{31}\sqrt{5}-\frac{14}{31}\sqrt{3}+\frac{8}{31}\sqrt{15}$$

Subtract a from both sides.

$$b\sqrt{3}=-\frac{35}{31}+\frac{20}{31}\sqrt{5}-\frac{14}{31}\sqrt{3}+\frac{8}{31}\sqrt{15}-a$$

The equation is in standard form.

$$\sqrt{3}b=-a+\frac{8\sqrt{15}}{31}+\frac{20\sqrt{5}}{31}-\frac{14\sqrt{3}}{31}-\frac{35}{31}$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+\frac{8\sqrt{15}}{31}+\frac{20\sqrt{5}}{31}-\frac{14\sqrt{3}}{31}-\frac{35}{31}}{\sqrt{3}}$$

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

$$b=\frac{-a+\frac{8\sqrt{15}}{31}+\frac{20\sqrt{5}}{31}-\frac{14\sqrt{3}}{31}-\frac{35}{31}}{\sqrt{3}}$$

Divide \frac{20\sqrt{5}}{31}-a+\frac{8\sqrt{15}}{31}-\frac{35}{31}-\frac{14\sqrt{3}}{31} by \sqrt{3}.

$$b=\frac{\sqrt{3}\left(-31a+8\sqrt{15}+20\sqrt{5}-14\sqrt{3}-35\right)}{93}$$

Solve for a

$$a=-\frac{\left(7-4\sqrt{5}\right)\left(-4\sqrt{15}b-7\sqrt{3}b+2\sqrt{3}+5\right)}{31}$$