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

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

Solve for b

$$b=-\frac{\sqrt{3}a}{3}+8\sqrt{3}+40$$

Steps for Solving Linear Equation

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

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

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}.

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

Calculate 7 to the power of 2 and get 49.

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

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

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

Calculate -4 to the power of 2 and get 16.

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

The square of \sqrt{3} is 3.

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

Multiply 16 and 3 to get 48.

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

Subtract 48 from 49 to get 1.

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

Anything divided by one gives itself.

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

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

$$35+27\sqrt{3}+4\left(\sqrt{3}\right)^{2}-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$$

The square of \sqrt{3} is 3.

$$35+27\sqrt{3}+4\times 3-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$35+27\sqrt{3}+12-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$$

Add 35 and 12 to get 47.

$$47+27\sqrt{3}-\frac{5+\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$$

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

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=a+b\sqrt{3}$$

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}.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

Calculate 7 to the power of 2 and get 49.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

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

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

Calculate 4 to the power of 2 and get 16.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}=a+b\sqrt{3}$$

The square of \sqrt{3} is 3.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-16\times 3}=a+b\sqrt{3}$$

Multiply 16 and 3 to get 48.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{49-48}=a+b\sqrt{3}$$

Subtract 48 from 49 to get 1.

$$47+27\sqrt{3}-\frac{\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{1}=a+b\sqrt{3}$$

Anything divided by one gives itself.

$$47+27\sqrt{3}-\left(5+\sqrt{3}\right)\left(7-4\sqrt{3}\right)=a+b\sqrt{3}$$

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

$$47+27\sqrt{3}-\left(35-13\sqrt{3}-4\left(\sqrt{3}\right)^{2}\right)=a+b\sqrt{3}$$

The square of \sqrt{3} is 3.

$$47+27\sqrt{3}-\left(35-13\sqrt{3}-4\times 3\right)=a+b\sqrt{3}$$

Multiply -4 and 3 to get -12.

$$47+27\sqrt{3}-\left(35-13\sqrt{3}-12\right)=a+b\sqrt{3}$$

Subtract 12 from 35 to get 23.

$$47+27\sqrt{3}-\left(23-13\sqrt{3}\right)=a+b\sqrt{3}$$

To find the opposite of 23-13\sqrt{3}, find the opposite of each term.

$$47+27\sqrt{3}-23+13\sqrt{3}=a+b\sqrt{3}$$

Subtract 23 from 47 to get 24.

$$24+27\sqrt{3}+13\sqrt{3}=a+b\sqrt{3}$$

Combine 27\sqrt{3} and 13\sqrt{3} to get 40\sqrt{3}.

$$24+40\sqrt{3}=a+b\sqrt{3}$$

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

$$a+b\sqrt{3}=24+40\sqrt{3}$$

Subtract a from both sides.

$$b\sqrt{3}=24+40\sqrt{3}-a$$

The equation is in standard form.

$$\sqrt{3}b=-a+40\sqrt{3}+24$$

Divide both sides by \sqrt{3}.

$$\frac{\sqrt{3}b}{\sqrt{3}}=\frac{-a+40\sqrt{3}+24}{\sqrt{3}}$$

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

$$b=\frac{-a+40\sqrt{3}+24}{\sqrt{3}}$$

Divide -a+40\sqrt{3}+24 by \sqrt{3}.

$$b=\frac{\sqrt{3}\left(-a+40\sqrt{3}+24\right)}{3}$$

Solve for a

$$a=\sqrt{3}\left(40-b\right)+24$$