Solve frac{7+sqrt{5}}{7-sqrt{5}}+frac{7-sqrt{5}}{7+sqrt{5}}=a+7/5sqrt{11}b

$$\frac { 7 + \sqrt { 5 } } { 7 - \sqrt { 5 } } + \frac { 7 - \sqrt { 5 } } { 7 + \sqrt { 5 } } = a + \frac { 7 } { 5 } \sqrt { 11 } b$$

Solve for a

$$a=-\frac{7\sqrt{11}b}{5}+\frac{27}{11}$$

Short Solution Steps

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

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Consider \left(7-\sqrt{5}\right)\left(7+\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}.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Square 7. Square \sqrt{5}.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{49-5}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Subtract 5 from 49 to get 44.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Multiply 7+\sqrt{5} and 7+\sqrt{5} to get \left(7+\sqrt{5}\right)^{2}.

$$\frac{\left(7+\sqrt{5}\right)^{2}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+\sqrt{5}\right)^{2}.

$$\frac{49+14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

The square of \sqrt{5} is 5.

$$\frac{49+14\sqrt{5}+5}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Add 49 and 5 to get 54.

$$\frac{54+14\sqrt{5}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

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

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}=a+\frac{7}{5}\sqrt{11}b$$

Consider \left(7+\sqrt{5}\right)\left(7-\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}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}=a+\frac{7}{5}\sqrt{11}b$$

Square 7. Square \sqrt{5}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{49-5}=a+\frac{7}{5}\sqrt{11}b$$

Subtract 5 from 49 to get 44.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{44}=a+\frac{7}{5}\sqrt{11}b$$

Multiply 7-\sqrt{5} and 7-\sqrt{5} to get \left(7-\sqrt{5}\right)^{2}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)^{2}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{5}\right)^{2}.

$$\frac{54+14\sqrt{5}}{44}+\frac{49-14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}=a+\frac{7}{5}\sqrt{11}b$$

The square of \sqrt{5} is 5.

$$\frac{54+14\sqrt{5}}{44}+\frac{49-14\sqrt{5}+5}{44}=a+\frac{7}{5}\sqrt{11}b$$

Add 49 and 5 to get 54.

$$\frac{54+14\sqrt{5}}{44}+\frac{54-14\sqrt{5}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Since \frac{54+14\sqrt{5}}{44} and \frac{54-14\sqrt{5}}{44} have the same denominator, add them by adding their numerators.

$$\frac{54+14\sqrt{5}+54-14\sqrt{5}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Do the calculations in 54+14\sqrt{5}+54-14\sqrt{5}.

$$\frac{108}{44}=a+\frac{7}{5}\sqrt{11}b$$

Reduce the fraction \frac{108}{44} to lowest terms by extracting and canceling out 4.

$$\frac{27}{11}=a+\frac{7}{5}\sqrt{11}b$$

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

$$a+\frac{7}{5}\sqrt{11}b=\frac{27}{11}$$

Subtract \frac{7}{5}\sqrt{11}b from both sides.

$$a=\frac{27}{11}-\frac{7}{5}\sqrt{11}b$$

Solve for b

$$b=-\frac{5\sqrt{11}\left(11a-27\right)}{847}$$

Steps for Solving Linear Equation

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

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{\left(7-\sqrt{5}\right)\left(7+\sqrt{5}\right)}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Consider \left(7-\sqrt{5}\right)\left(7+\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}.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Square 7. Square \sqrt{5}.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{49-5}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Subtract 5 from 49 to get 44.

$$\frac{\left(7+\sqrt{5}\right)\left(7+\sqrt{5}\right)}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Multiply 7+\sqrt{5} and 7+\sqrt{5} to get \left(7+\sqrt{5}\right)^{2}.

$$\frac{\left(7+\sqrt{5}\right)^{2}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+\sqrt{5}\right)^{2}.

$$\frac{49+14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

The square of \sqrt{5} is 5.

$$\frac{49+14\sqrt{5}+5}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

Add 49 and 5 to get 54.

$$\frac{54+14\sqrt{5}}{44}+\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{5}\sqrt{11}b$$

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

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{\left(7+\sqrt{5}\right)\left(7-\sqrt{5}\right)}=a+\frac{7}{5}\sqrt{11}b$$

Consider \left(7+\sqrt{5}\right)\left(7-\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}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{7^{2}-\left(\sqrt{5}\right)^{2}}=a+\frac{7}{5}\sqrt{11}b$$

Square 7. Square \sqrt{5}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{49-5}=a+\frac{7}{5}\sqrt{11}b$$

Subtract 5 from 49 to get 44.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)\left(7-\sqrt{5}\right)}{44}=a+\frac{7}{5}\sqrt{11}b$$

Multiply 7-\sqrt{5} and 7-\sqrt{5} to get \left(7-\sqrt{5}\right)^{2}.

$$\frac{54+14\sqrt{5}}{44}+\frac{\left(7-\sqrt{5}\right)^{2}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-\sqrt{5}\right)^{2}.

$$\frac{54+14\sqrt{5}}{44}+\frac{49-14\sqrt{5}+\left(\sqrt{5}\right)^{2}}{44}=a+\frac{7}{5}\sqrt{11}b$$

The square of \sqrt{5} is 5.

$$\frac{54+14\sqrt{5}}{44}+\frac{49-14\sqrt{5}+5}{44}=a+\frac{7}{5}\sqrt{11}b$$

Add 49 and 5 to get 54.

$$\frac{54+14\sqrt{5}}{44}+\frac{54-14\sqrt{5}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Since \frac{54+14\sqrt{5}}{44} and \frac{54-14\sqrt{5}}{44} have the same denominator, add them by adding their numerators.

$$\frac{54+14\sqrt{5}+54-14\sqrt{5}}{44}=a+\frac{7}{5}\sqrt{11}b$$

Do the calculations in 54+14\sqrt{5}+54-14\sqrt{5}.

$$\frac{108}{44}=a+\frac{7}{5}\sqrt{11}b$$

Reduce the fraction \frac{108}{44} to lowest terms by extracting and canceling out 4.

$$\frac{27}{11}=a+\frac{7}{5}\sqrt{11}b$$

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

$$a+\frac{7}{5}\sqrt{11}b=\frac{27}{11}$$

Subtract a from both sides.

$$\frac{7}{5}\sqrt{11}b=\frac{27}{11}-a$$

The equation is in standard form.

$$\frac{7\sqrt{11}}{5}b=\frac{27}{11}-a$$

Divide both sides by \frac{7}{5}\sqrt{11}.

$$\frac{5\times \frac{7\sqrt{11}}{5}b}{7\sqrt{11}}=\frac{5\left(\frac{27}{11}-a\right)}{7\sqrt{11}}$$

Dividing by \frac{7}{5}\sqrt{11} undoes the multiplication by \frac{7}{5}\sqrt{11}.

$$b=\frac{5\left(\frac{27}{11}-a\right)}{7\sqrt{11}}$$

Divide \frac{27}{11}-a by \frac{7}{5}\sqrt{11}.

$$b=\frac{5\sqrt{11}\left(27-11a\right)}{847}$$