Elementary Algebra

{$\frac{7-3\sqrt {5}}{2},\frac{7+3\sqrt {5}}{2}$}
First, we add the fractions on the left hand side by taking their LCM. Upon inspection, the LCM is found to be $n$: $n+\frac{1}{n}=7$ $\frac{n(n)+1(1)}{n}=7$ $\frac{n^{2}+1}{n}=7$ $\frac{n^{2}+1}{n}=\frac{7}{1}$ Now, we cross multiply the two fractions in order to create a quadratic equation: $\frac{n^{2}+1}{n}=\frac{7}{1}$ $n^{2}+1=7n$ $n^{2}-7n+1=0$ Now, we will use the quadratic formula to solve the equation: Step 1: Comparing $n^{2}-7n+1=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$; $a=1$, $b=-7$ and $c=1$ Step 2: The quadratic formula is: $n=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $n=\frac{-(-7) \pm \sqrt {(-7)^{2}-4(1)(1)}}{2(1)}$ Step 4: $n=\frac{7 \pm \sqrt {49-4}}{2}$ Step 5: $n=\frac{7 \pm \sqrt {45}}{2}$ Step 6: $n=\frac{7 \pm \sqrt {9\times5}}{2}$ Step 7: $n=\frac{7 \pm 3\sqrt {5}}{2}$ Step 8: $n=\frac{7-3\sqrt {5}}{2}$ or $n=\frac{7+3\sqrt {5}}{2}$ Step 9: Therefore, the solution set is {$\frac{7-3\sqrt {5}}{2},\frac{7+3\sqrt {5}}{2}$}.