Solve G=frac{1-3sqrt{5}}{1+3sqrt{5}},F=frac{sqrt{52}}{sqrt{2}}

$$G = \frac { 1 - 3 \sqrt { 5 } } { 1 + 3 \sqrt { 5 } } , F = \frac { \sqrt { 52 } } { \sqrt { 2 } }$$

Solve for G, F

$$G=\frac{3\sqrt{5}-23}{22}\approx -0.740536185$$

$$F=\sqrt{26}\approx 5.099019514$$

Solution Steps

Consider the first equation. Rationalize the denominator of \frac{1-3\sqrt{5}}{1+3\sqrt{5}} by multiplying numerator and denominator by 1-3\sqrt{5}.

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

Consider \left(1+3\sqrt{5}\right)\left(1-3\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}.

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

Multiply 1-3\sqrt{5} and 1-3\sqrt{5} to get \left(1-3\sqrt{5}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-3\sqrt{5}\right)^{2}.

$$G=\frac{1-6\sqrt{5}+9\left(\sqrt{5}\right)^{2}}{1^{2}-\left(3\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$G=\frac{1-6\sqrt{5}+9\times 5}{1^{2}-\left(3\sqrt{5}\right)^{2}}$$

Multiply 9 and 5 to get 45.

$$G=\frac{1-6\sqrt{5}+45}{1^{2}-\left(3\sqrt{5}\right)^{2}}$$

Add 1 and 45 to get 46.

$$G=\frac{46-6\sqrt{5}}{1^{2}-\left(3\sqrt{5}\right)^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$G=\frac{46-6\sqrt{5}}{1-\left(3\sqrt{5}\right)^{2}}$$

Expand \left(3\sqrt{5}\right)^{2}.

$$G=\frac{46-6\sqrt{5}}{1-3^{2}\left(\sqrt{5}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$G=\frac{46-6\sqrt{5}}{1-9\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$G=\frac{46-6\sqrt{5}}{1-9\times 5}$$

Multiply 9 and 5 to get 45.

$$G=\frac{46-6\sqrt{5}}{1-45}$$

Subtract 45 from 1 to get -44.

$$G=\frac{46-6\sqrt{5}}{-44}$$

Divide each term of 46-6\sqrt{5} by -44 to get -\frac{23}{22}+\frac{3}{22}\sqrt{5}.

$$G=-\frac{23}{22}+\frac{3}{22}\sqrt{5}$$

Consider the second equation. Rewrite the division of square roots \frac{\sqrt{52}}{\sqrt{2}} as the square root of the division \sqrt{\frac{52}{2}} and perform the division.

$$F=\sqrt{26}$$

The system is now solved.

$$G=-\frac{23}{22}+\frac{3}{22}\sqrt{5} F=\sqrt{26}$$