Question

1. the sum of a number, x, and its reciprocal, is $\frac{39}{10}$. form an equation and find the original number.

Explanation:

Step1: Let the number be \(x\).

The reciprocal of the number is \(\frac{1}{x}\). The sum of the number and its reciprocal is \(x+\frac{1}{x}\), and we know it is equal to \(\frac{37}{6}\). So the equation is \(x + \frac{1}{x}=\frac{37}{6}\).

Step2: Multiply through by \(6x\) to clear the fractions.

\[6x\times x+6x\times\frac{1}{x}=6x\times\frac{37}{6}\]
\[6x^{2}+ 6 = 37x\]
\[6x^{2}-37x + 6=0\]

Step3: Solve the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 6\), \(b=-37\), \(c = 6\)) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or by factoring.

Factor the quadratic: \(6x^{2}-37x + 6=(6x - 1)(x - 6)=0\)
Setting each factor equal to zero gives \(6x-1 = 0\) or \(x - 6=0\).
If \(6x-1=0\), then \(x=\frac{1}{6}\); if \(x - 6=0\), then \(x = 6\).

Answer:

The number is \(6\) or \(\frac{1}{6}\)